Rd Sharma Xi 2019 for Class 12 Commerce Math Chapter 9 - Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle

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Rd Sharma Xi 2019 Solutions for Class 12 Commerce Math Chapter 9 Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle are provided here with simple step-by-step explanations. These solutions for Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle are extremely popular among Class 12 Commerce students for Math Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2019 Book of Class 12 Commerce Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2019 Solutions. All Rd Sharma Xi 2019 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 9.28:

Question 1:

Prove that:
$\sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} = \tan x$

Answer:

$LHS = \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} = \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x}} [∵ 1 - \cos 2 x = 2 \sin^{2} x and 1 + \cos 2 x = 2 \cos^{2} x] = \frac{\sin x}{\cos x} = \tan x = RHS Hence proved .$

Page No 9.28:

Question 2:

Prove that:
$\frac{\sin 2 x}{1 - \cos 2 x} = c o t x$

Answer:

$LHS = \frac{\sin 2 x}{1 - \cos 2 x} = \frac{2 \sin x \times \cos x}{2 \sin^{2} x} (∵ \sin 2 x, 1 - \cos 2 x = 2 \sin^{2} x) = \frac{2 \sin x \times \cos x}{2 \sin x \times \sin x} = \frac{\cos x}{\sin x} = \cot x = RHS Hence proved .$

Page No 9.28:

Question 3:

Prove that:
$\frac{\sin 2 x}{1 + \cos 2 x} = \tan x$

Answer:

$LHS = \frac{\sin 2 x}{1 + \cos 2 x} = \frac{2 \sin x \times \cos x}{1 + 2 \cos^{2} x - 1} [∵ \sin 2 x = 2 \sin x \times \cos x and \cos 2 x = 2 \cos^{2} x - 1] = \frac{2 \sin x \times \cos x}{2 \cos x \times \cos x} = \frac{\sin x}{\cos x} = \tan x = RHS Hence proved .$

Page No 9.28:

Question 4:

Prove that:
$\sqrt{2 + \sqrt{2 + 2 \cos 4 x}} = 2 \cos x$

Answer:

$LHS = \sqrt{2 + \sqrt{2 + 2 \cos 4 x}} = \sqrt{2 + \sqrt{2 (1 + \cos 4 x)}} = \sqrt{2 + \sqrt{2 \times 2 \cos^{2} 2 x}} (∵ 2 \cos^{2} 2 x = 1 + \cos 4 x) = \sqrt{2 + 2 \cos 2 x}$

$= \sqrt{2 (1 + \cos 2 x)} = \sqrt{2.2 \cos^{2} x} (∵ 2 \cos^{2} x = 1 + \cos 2 x) = 2 \cos x = RHS Hence proved .$

Page No 9.28:

Question 5:

Prove that:
$\frac{1 - \cos 2 x + \sin 2 x}{1 + \cos 2 x + \sin 2 x} = \tan x$

Answer:

$LHS = \frac{1 - \cos 2 x + \sin 2 x}{1 + \cos 2 x + \sin 2 x}$

$= \frac{2 \sin^{2} x + \sin 2 x}{2 \cos^{2} x + \sin 2 x} [∵ 2 \sin^{2} x = 1 - \cos 2 x and 2 \cos^{2} x = 1 + \cos 2 x] = \frac{2 \sin^{2} x + 2 \sin x \cos x}{2 \cos^{2} x + 2 \sin x \cos x} (∵ \sin 2 x = 2 \sin x \cos x) = \frac{2 \sin x (\sin x + \cos x)}{2 \cos x (\cos x + \sin x)} = tanθ = RHS Hence proved .$

Page No 9.28:

Question 6:

Prove that:
$\frac{\sin x + \sin 2 x}{1 + \cos x + \cos 2 x} = \tan x$

Answer:

$LHS = \frac{\sin x + \sin 2 x}{1 + \cos x + \cos 2 x} = \frac{\sin x + \sin 2 x}{\cos x + (1 + \cos 2 x)}$

$= \frac{\sin x + 2 \sin x \cos x}{\cos x + 2 \cos^{2} x} [∵ \sin 2 x = 2 \sin x \cos x and 2 \cos^{2} x = 1 + \cos 2 x]$

$= \frac{\sin x (1 + 2 \cos x)}{\cos x (1 + 2 \cos x)} = \tan x = RHS Hence proved .$

Page No 9.28:

Question 7:

Prove that:
$\frac{\cos 2 x}{1 + \sin 2 x} = \tan (\frac{π}{4} - x)$

Answer:

$LHS = \frac{\cos 2 x}{1 + \sin 2 x}$

$= \frac{\cos^{2} x - \sin^{2} x}{\sin^{2} x + \cos^{2} x + 2 \sin x \times \cos x} [∵ \cos 2 x = \cos^{2} x - \sin^{2} x and \sin^{2} x + \cos^{2} x = 1] = \frac{(\cos x - \sin x) (\cos x + \sin x)}{{(\cos x + \sin x)}^{2}} [∵ a^{2} - b^{2} = (a + b) (a - b)] = \frac{\cos x - \sin x}{\cos x + \sin x}$

On dividing the numerator and denominator by cos x, we get
$= \frac{1 - \tan x}{1 + \tan x} = \tan (\frac{π}{4} - x) = RHS Hence proved .$

Page No 9.28:

Question 8:

Prove that:
$\frac{\cos x}{1 - \sin x} = \tan (\frac{π}{4} + \frac{x}{2})$

Answer:

$LHS = \frac{\cos x}{1 - \sin x}$
$= \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \times \cos \frac{x}{2}} [∵ \cos x = \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}, \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} and \sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} = 1] = \frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) (\cos \frac{x}{2} + \sin \frac{x}{2})}{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$

On dividing the numerator and denominator by $\cos \frac{x}{2}$ , we get

$= \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} = \tan (\frac{π}{4} + \frac{x}{2}) = RHS Hence proved .$

Page No 9.28:

Question 9:

Prove that:

$\cos^{2} \frac{π}{8} + \cos^{2} \frac{3 π}{8} + \cos^{2} \frac{5 π}{8} + \cos^{2} \frac{7 π}{8} = 2$

Answer:

$LHS = \cos^{2} \frac{π}{8} + \cos^{2} \frac{3 π}{8} + \cos^{2} \frac{5 π}{8} + \cos^{2} \frac{7 π}{8} = \cos^{2} \frac{π}{8} + \cos^{2} \frac{3 π}{8} + \cos^{2} (π - \frac{3 π}{8}) + \cos^{2} (π - \frac{π}{8}) = \cos^{2} \frac{π}{8} + \cos^{2} \frac{3 π}{8} + {\{- \cos (\frac{3 π}{8})\}}^{2} + {\{- \cos (\frac{π}{8})\}}^{2}$

$= \cos^{2} \frac{π}{8} + \cos^{2} \frac{3 π}{8} + \cos^{2} \frac{3 π}{8} + \cos^{2} \frac{π}{8} = 2 (\cos^{2} \frac{π}{8} + \cos^{2} \frac{3 π}{8}) = 2 \{\cos^{2} \frac{π}{8} + \cos^{2} (\frac{π}{2} - \frac{π}{8})\} = 2 (\cos^{2} \frac{π}{8} + \sin^{2} \frac{π}{8}) = 2 = RHS Hence proved .$

Page No 9.28:

Question 10:

Prove that:

$\sin^{2} \frac{π}{8} + \sin^{2} \frac{3 π}{8} + \sin^{2} \frac{5 π}{8} + \sin^{2} \frac{7 π}{8} = 2$

Answer:

$LHS = \sin^{2} \frac{π}{8} + \sin^{2} \frac{3 π}{8} + \sin^{2} \frac{5 π}{8} + \sin^{2} \frac{7 π}{8} = \sin^{2} (\frac{π}{2} - \frac{3 π}{8}) + \sin^{2} (\frac{π}{2} - \frac{π}{8}) + \sin^{2} \frac{5 π}{8} + \sin^{2} \frac{7 π}{8} = \cos^{2} \frac{3 π}{8} + \sin^{2} \frac{π}{8} + \sin^{2} (π - \frac{3 π}{8}) + \sin^{2} (π - \frac{π}{8})$

$= \cos^{2} \frac{3 π}{8} + \sin^{2} \frac{π}{8} + \sin^{2} \frac{3 π}{8} + \cos^{2} \frac{π}{8} = (\cos^{2} \frac{π}{8} + \sin^{2} \frac{π}{8}) + (\cos^{2} \frac{3 π}{8} + \sin^{2} \frac{3 π}{8}) = 1 + 1 = 2 = RHS Hence proved .$

Page No 9.28:

Question 11:

Prove that:
$(\cos α + \cos β^{2}) + {(\sin α + \sin β)}^{2} = 4 \cos^{2} (\frac{α - β}{2})$

Answer:

$LHS = {(cosα + cosβ)}^{2} + {(sinα + sinβ)}^{2} = \cos^{2} α + \cos^{2} β + 2 cosαcosβ + \sin^{2} α + \sin^{2} β + 2 sinαsinβ = (\cos^{2} α + \sin^{2} α) + (\cos^{2} β + \sin^{2} β) + 2 (cosαcosβ + sinαsinβ) = 1 + 1 + 2 \cos (α - β) = 2 \{1 + \cos (α - β)\} = 2 \{2 \cos^{2} (\frac{α - β}{2})\} = 4 \cos^{2} (\frac{α - β}{2}) = RHS Hence proved .$

Page No 9.28:

Question 12:

Prove that:
$\sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{\sqrt{2}} \sin x$

Answer:

$LHS = \sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{2} \{1 - \cos 2 (\frac{π}{8} + \frac{x}{2})\} - \frac{1}{2} \{1 - \cos 2 (\frac{π}{8} - \frac{x}{2})\} = \frac{1}{2} \{\cos (\frac{π}{4} - x) - \cos (\frac{π}{4} + x)\}$

Using the identity $\cos C - \cos D = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}$ , we get

$= \frac{1}{2} \{- 2 \sin (\frac{(\frac{π}{4} - x) + (\frac{π}{4} + x)}{2}) \sin (\frac{(\frac{π}{4} - x) - (\frac{π}{4} + x)}{2})\} = - \sin \frac{π}{4} \sin (- x) = \sin \frac{π}{4} \sin x [∵ \sin (- x) = - \sin x] = \frac{1}{\sqrt{2}} \sin x = RHS Hence proved .$

Page No 9.28:

Question 13:

Prove that:
$1 + \cos^{2} 2 x = 2 (\cos^{4} x + \sin^{4} x)$

Answer:

$LHS = 1 + \cos^{2} 2 x$
Using the identity $\cos 2 x = \cos^{2} x - s {in}^{2} x$ , we get

$LHS = 1 + {(\cos^{2} x - \sin^{2} x)}^{2} = 1 + \cos^{4} x + \sin^{4} x - 2 \cos^{2} x \sin^{2} x$

$= {(\cos^{2} x + \sin^{2} x)}^{2} + \cos^{4} x + \sin^{4} x - 2 \cos^{2} x \sin^{2} x [∵ \cos^{2} x + \sin^{2} x = 1] = \cos^{4} x + \sin^{4} x + 2 \cos^{2} x \sin^{2} x + \cos^{4} x + \sin^{4} x - 2 \cos^{2} x \sin^{2} x = 2 (\cos^{4} x + \sin^{4} x) = RHS Hence proved .$

Page No 9.28:

Question 14:

Prove that:
$\cos^{3} 2 x + 3 \cos 2 x = 4 (\cos^{6} x - \sin^{6} x)$

Answer:

$RHS = 4 (\cos^{6} x - \sin^{6} x) = 4 [{(\cos^{2} x)}^{3} - {(\sin^{2} x)}^{3}]$

Using the identity $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ , we get

$= 4 (\cos^{2} x - \sin^{2} x) (\cos^{4} x + \sin^{4} x + \sin^{2} x \cos^{2} x) = 4 (\cos^{2} x - \sin^{2} x) (\cos^{4} x + \sin^{4} x + \sin^{2} x \cos^{2} x)$

$= 4 \cos 2 x [{(\cos^{2} x - \sin^{2} x)}^{2} + 2 \sin^{2} x \cos^{2} x + \sin^{2} x \cos^{2} x] = 4 \cos 2 x [\cos^{2} 2 x + 3 \sin^{2} x \cos^{2} x] = 4 \cos 2 x [\cos^{2} 2 x + 3 (\frac{1 - \cos^{2} x}{2}) (\frac{1 + \cos^{2} x}{2})]$
$= 4 \cos 2 x [\cos^{2} 2 x + \frac{3}{4} (1 - \cos^{2} 2 x)] = \cos 2 x [4 \cos^{2} 2 x + 3 (1 - \cos^{2} 2 x)] = \cos 2 x [4 \cos^{2} 2 x + 3 - 3 \cos^{2} 2 x] = \cos 2 x [\cos^{2} 2 x + 3] = \cos^{3} 2 x + 3 \cos 2 x = LHS Hence proved .$

Page No 9.28:

Question 15:

Prove that:
$(\sin 3 x + \sin x) \sin x + (\cos 3 x - \cos x) \cos x = 0$

Answer:

$LHS = (\sin 3 x + \sin x) \sin x + (\cos 3 x - \cos x) \cos x$

Using the identities $\sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2} and \cos C - \cos D = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}$ , we get

$LHS = (2 \sin \frac{3 x + x}{2} \times \cos \frac{3 x - x}{2} \times \sin x) + (- 2 \sin \frac{3 x + x}{2} \times \sin \frac{3 x - x}{2}) \cos x = (2 \sin 2 x \times \cos x \times \sin x) - (2 \sin 2 x \times \sin x \cos x) = 0 = RHS Hence proved .$

Page No 9.28:

Question 16:

Prove that:

$\cos^{2} (\frac{π}{4} - x) - \sin^{2} (\frac{π}{4} - x) = \sin 2 x$

Answer:

$LHS = \cos^{2} (\frac{π}{4} - x) - \sin^{2} (\frac{π}{4} - x) = \cos 2 (\frac{π}{4} - x) [∵ \cos^{2} α - \sin^{2} α = \cos 2 α] = \cos (\frac{π}{2} - 2 x) = \sin 2 x = RHS [∵ \cos (\frac{π}{2} - 2 α) = \sin 2 α] Hence proved .$

Page No 9.28:

Question 17:

Prove that:
$\cos 4 x = 1 - 8 \cos^{2} x + 8 \cos^{4} x$

Answer:

$LHS = \cos 4 x = \cos (2 \times 2 x) = 2 \cos^{2} \times 2 x - 1 [∵ \cos 2 θ = 2 \cos^{2} θ - 1] = 2 {(2 \cos^{2} x - 1)}^{2} - 1 [∵ \cos^{2} 2 θ = {(2 \cos^{2} θ - 1)}^{2}]$

$= 2 (4 \cos^{4} x - 4 \cos^{2} x + 1) - 1 = 8 \cos^{4} x - 8 \cos^{2} x + 1 = 1 - 8 \cos^{2} x + 8 \cos^{4} x = RHS Hence proved .$

Page No 9.28:

Question 18:

Prove that:
$\sin 4 x = 4 \sin x \cos^{3} x - 4 \cos x \sin^{3} x$

Answer:

$LHS = sin 4 x = 2 \sin 2 x \cos 2 x (∵ \sin 2 θ = 2 sinθ cosθ)$

Now, using the identities $\sin 2 α = 2 \sin α \cos α and \cos 2 α = \cos^{2} α - \sin^{2} α$ , we get

$LHS = 2 (2 \sin x \cos x) . (\cos^{2} x - \sin^{2} x) = 4 \sin x \cos^{3} x - 4 \sin^{3} x \cos x = RHS Hence proved .$

Page No 9.28:

Question 19:

Show that: $3 {(\sin x - \cos x)}^{4} + 6 {(\sin x + \cos)}^{2} + 4 (\sin^{6} x + \cos^{6} x) = 13$

Answer:

$LHS = 3 {(\sin x - \cos x)}^{4} + 6 {(\sin x + \cos x)}^{2} + 4 (\sin^{6} x + \cos^{6} x) = 3 {\{{(\sin x - \cos x)}^{2}\}}^{2} + 6 (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x) + 4 \{{(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}\}$

$= 3 (\sin^{2} x + \cos^{2} x - 2 \sin x \cos x)^{2} + 6 (1 + \sin 2 x) + 4 (\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x) = 3 (1 - \sin 2 x)^{2} + 6 + 6 \sin 2 x + 4 (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x) = 3 {(1 - \sin 2 x)}^{2} + 6 + 6 \sin 2 x + 4 \{(\sin^{2} x)^{2} + (\cos^{2} x)^{2}\} - 12 \sin^{2} x \cos^{2} x$

$= 3 {(1 - \sin 2 x)}^{2} + 6 + 6 s in 2 x + 4 \{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x\} - 12 \sin^{2} x \cos^{2} x = 3 (1 - \sin 2 x)^{2} + 6 + 6 \sin 2 x + 4 \{1 - 2 \sin^{2} x \cos^{2} x\} - 12 \sin^{2} x \cos^{2} x = 3 (1 - 2 \sin 2 x + \sin^{2} 2 x) + 6 + 6 \sin 2 x + 4 - 20 \sin^{2} x \cos^{2} x = - 6 \sin 2 x + 3 \sin^{2} 2 x + 13 + 6 \sin 2 x - 5 {(2 \sin x \cos x)}^{2} = 3 \sin^{2} 2 x + 13 - 5 \sin^{2} 2 x = 13 - 2 \sin^{2} 2 x = RHS Hence proved .$

Page No 9.28:

Question 20:

Show that: $2 (\sin^{6} x + \cos^{6} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 = 0$

Answer:

$LHS = 2 (\sin^{6} x + \cos^{6} x) -3 (\sin^{4} x + \cos^{4} x) + 1 = 2 \{{(\sin^{2} x)}^{3} + {(\cos^{2})}^{3}\} - 3 (\sin^{4} x + \cos^{4} x) + 1$

$= 2 [(\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x)] - 3 (\sin^{4} x + \cos^{4} x) + 1 = 2 (\sin^{4} x + \cos^{4} x) - 2 \sin^{2} x \cos^{2} x - 3 (\sin^{4} x + \cos^{4} x) + 1 = - [\sin^{4} x + \cos^{4} x + 2 \sin^{2} x \cos^{2} x] + 1 = - 1 + 1 = 0 = RHS Hence proved .$

Page No 9.28:

Question 21:

Prove that: $\cos^{6} A - \sin^{6} A = \cos 2 A (1 - \frac{1}{4} \sin^{2} 2 A)$

Answer:

$LHS = \cos^{6} A - \sin^{6} A = {(\cos^{2} A)}^{3} - {(\sin^{2} A)}^{3} = (\cos^{2} A - \sin^{2} A) (\cos^{4} A + \sin^{2} A . \cos^{2} A + \sin^{4} A)$

$= \cos 2 A (\cos^{4} A + 2 \sin^{2} A \cos^{2} A + \sin^{4} A - \sin^{2} A \cos^{2} A) = \cos 2 A \{{(\sin^{2} A + \cos^{2} A)}^{2} - \frac{1}{4} \times 4 \sin^{2} A \cos^{2} A\}$

$= \cos 2 A \{{(\sin^{2} A + \cos^{2} A)}^{2} - \frac{1}{4} {(2 \sin A \cos A)}^{2}\} = \cos 2 A \{1 - \frac{1}{4} {(\sin 2 A)}^{2}\} = \cos 2 A \{1 - \frac{1}{4} \sin^{2} 2 A\} = RHS Hence proved .$

Page No 9.28:

Question 22:

Prove that: $\tan (\frac{π}{4} + x) + \tan (\frac{π}{4} - x) = 2 \sec 2 x$

Answer:

$LHS = \tan (\frac{π}{4} + x) + \tan (\frac{π}{4} - x) = \frac{\tan \frac{π}{4} + \tan x}{1 - \tan \frac{π}{4} \tan x} + \frac{\tan \frac{π}{4} - \tan x}{1 + \tan \frac{π}{4} \tan x} [∵ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} and \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}]$

$= \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} = \frac{{(1 + \tan x)}^{2} + {(1 - \tan x)}^{2}}{(1 + \tan x) (1 - \tan x)} = \frac{2 (1 + \tan^{2} x)}{(1 - \tan^{2} x)} = \frac{2 (\sec^{2} x)}{1 - \frac{\sin^{2} x}{\cos^{2} x}}$

$= \frac{2 (\sec^{2} x) (\cos^{2} x)}{\cos 2 x} (∵ \cos^{2} x - \sin^{2} x = \cos 2 x) = \frac{2 \times 1}{\cos 2 x} = 2 \sec 2 x = RHS Hence proved .$

Page No 9.28:

Question 23:

Prove that: $\cot^{2} x - \tan^{2} x = 4 \cot 2 x cosec 2 x$

Answer:

$LHS = \cot^{2} x - \tan^{2} x = \frac{\cos^{2} x}{\sin^{2} x} - \frac{\sin^{2} x}{\cos^{2} x} = \frac{{(\cos^{2} x)}^{2} - {(\sin^{2} x)}^{2}}{\sin^{2} x \cos^{2} x}$

$= \frac{(\cos^{2} x + \sin^{2} x) (\cos^{2} x - \sin^{2} x)}{\sin^{2} x \cos^{2} x} = \frac{1 \times (\cos 2 x)}{\sin^{2} x \cos^{2} x} (∵ \cos^{2} x - \sin^{2} x = \cos 2 x) = \frac{4 \cos 2 x}{4 \sin^{2} x \cos^{2} x}$

$= \frac{4 (\cos 2 x)}{{(\sin 2 x)}^{2}} = \frac{4 (\cos 2 x)}{(\sin 2 x)} \times \frac{1}{(\sin 2 x)} = 4 \cot 2 x cosec 2 x = RHS Hence proved .$

Page No 9.28:

Question 24:

Prove that:
$\cos 4 x - \cos 4 α = 8 (\cos x - \cos α) (\cos x + \cos α) (\cos x - \sin α) (\cos x + \sin α)$

Answer:

$RHS = 8 (\cos x - cosα) (\cos x + cosα) (\cos x - sinα) (\cos x + sinα) = 8 (\cos^{2} x - \cos^{2} α) (\cos^{2} x - \sin^{2} α) = 8 (\cos^{4} x - \cos^{2} x \times \sin^{2} α - \cos^{2} α \times \cos^{2} x + \cos^{2} α \times \sin^{2} α) = 8 \{\cos^{4} x - \cos^{2} x (\sin^{2} α + \cos^{2} α) + \cos^{2} α \times \sin^{2} α\} = 8 \{\cos^{4} x - \cos^{2} x + \cos^{2} α \times (1 - \cos^{2} α)\} = 8 \{\cos^{4} x - \cos^{2} x + \cos^{2} α - \cos^{4} α\} = 8 \{\cos^{2} x (\cos^{2} x - 1) + \cos^{2} α \times (1 - \cos^{2} α)\}$

$= 8 \{\frac{1}{2} \cos^{2} x (2 \cos^{2} x - 2) + \frac{1}{2} \cos^{2} α \times (2 - 2 \cos^{2} α)\} = 8 \{\frac{1}{2} \cos^{2} x (2 \cos^{2} x - 1 - 1) - \frac{1}{2} \cos^{2} α \times (2 \cos^{2} α - 1 - 1)\} = 8 \{\frac{1}{2} \cos^{2} x (\cos 2 x - 1) - \frac{1}{2} \cos^{2} α \times (\cos 2 α - 1)\} (∵ \cos 2 α = 2 \cos^{2} α - 1) = 8 [\frac{1}{4} \{2 \cos^{2} x (\cos 2 x - 1) - 2 \cos^{2} α \times (\cos 2 α - 1)\}] = 8 [\frac{1}{4} \{(1 + \cos 2 x) (\cos 2 x - 1) - (1 + \cos 2 α) (\cos 2 α - 1)\}]$

$= 8 [\frac{1}{4} \{\cos^{2} 2 x - 1 - \cos^{2} 2 α + 1\}] = 8 [\frac{1}{8} \{2 \cos^{2} 2 x - 2 \cos^{2} 2 α\}] = [\{(1 + \cos 4 x) - (1 + \cos 4 α)\}] = [1 + \cos 4 x - 1 - \cos 4 α] = \cos 4 x - \cos 4 α = LHS Hence proved .$

Page No 9.28:

Question 25:

Prove that $\sin 3 x + \sin 2 x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$

Answer:

$LHS = \sin 3 x + \sin 2 x - \sin x = \sin 3 x + 2 \sin (\frac{2 x - x}{2}) \cos (\frac{2 x + x}{2}) [∵ \sin A - \sin B = 2 \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2})] = \sin 3 x + 2 \sin (\frac{x}{2}) \cos (\frac{3 x}{2}) = 2 \sin (\frac{3 x}{2}) \cos (\frac{3 x}{2}) + 2 \sin (\frac{3 x}{2}) \cos \frac{x}{2} [∵ \sin 2 A = 2 \sin A \cos A] = 2 \cos (\frac{3 x}{2}) [\sin (\frac{3 x}{2}) \cos (\frac{x}{2})] = 2 \cos (\frac{3 x}{2}) [2 \sin (\frac{\frac{3 x}{2} + \frac{x}{2}}{2}) \cos (\frac{\frac{3 x}{2} - \frac{x}{2}}{2})] [∵ \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})] = 2 \cos (\frac{3 x}{2}) [2 \sin (x) \cos (\frac{x}{2})] = 4 \sin (x) \cos (\frac{x}{2}) \cos (\frac{3 x}{2}) = RHS Hence proved .$

Page No 9.29:

Question 26:

$\tan 82 \frac{1 °}{2} = (\sqrt{3} + \sqrt{2}) (\sqrt{2} + 1) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$

Answer:

$Here, \tan (82.5) ° = \tan (90 - 7.5) ° = \cot (7.5) ° = \frac{1}{\tan (7.5) °} We know, \tan (\frac{x}{2}) = \frac{\sin x}{1 + \cos x} On putting x = 15 °, we get \tan {(\frac{15}{2})}^{°} = \frac{\sin 15 °}{1 + \cos 15 °} = \frac{\sin (45 - 30) °}{1 + \cos (45 - 30) °} = \frac{\sin 45 ° \cos 30 ° - \sin 30 ° \cos 45 °}{1 + \cos 45 ° \cos 30 ° + \sin 45 ° \sin 30 °} = \frac{(\frac{1}{\sqrt{2}}) \times (\frac{\sqrt{3}}{2}) - (\frac{1}{2}) \times (\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} \times \frac{1}{2})} = \frac{\frac{\sqrt{3}}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}}}{1 + \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}} = \frac{\sqrt{3} - 1}{2 \sqrt{2} + \sqrt{3} + 1} Now, \tan (82.5) ° = \frac{1}{\tan (7.5) °} = \frac{2 \sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} = \frac{2 \sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{\sqrt{3} + 1 (2 \sqrt{2} + \sqrt{3} + 1)}{{(\sqrt{3})}^{2} - 1^{2}} = \frac{2 \sqrt{6} + 3 + \sqrt{3} + 2 \sqrt{2} + \sqrt{3} + 1}{3 - 1} = \frac{2 \sqrt{6} + 2 \sqrt{3} + 2 \sqrt{2} + 4}{2} = \sqrt{6} + \sqrt{3} + \sqrt{2} + 2 = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} . . . (1) = \sqrt{6} + \sqrt{3} + 2 + \sqrt{2} = \sqrt{3} (\sqrt{2} + 1) + \sqrt{2} (\sqrt{2} + 1) = (\sqrt{3} + \sqrt{2}) (\sqrt{2} + 1) . . . (2) From eqs . (1) and (2), we get \tan (82.5) ° = (\sqrt{3} + \sqrt{2}) (\sqrt{2} + 1) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$

Page No 9.29:

Question 27:

Prove that: $\cot \frac{π}{8} = \sqrt{2} + 1$

Answer:

$\frac{π}{8} = {(22 \frac{1}{2})}^{°} Let A = {(22 \frac{1}{2})}^{°} Using the identity \cot 2 A = \frac{\cot^{2} A - 1}{2 \cot A}, we get \cot 45 ° = \frac{\cot^{2} {(22 \frac{1}{2})}^{°} - 1}{2 \cot {(22 \frac{1}{2})}^{°}} \Rightarrow 1 = \frac{\cot^{2} {(22 \frac{1}{2})}^{°} - 1}{2 \cot {(22 \frac{1}{2})}^{°}} (∵ \cot 45 ° = 1) \Rightarrow 2 \cot {(22 \frac{1}{2})}^{°} - \cot^{2} {(22 \frac{1}{2})}^{°} + 1 = 0$

$\Rightarrow \cot^{2} {(22 \frac{1}{2})}^{\circ} - 2 \cot {(22 \frac{1}{2})}^{\circ} - 1 = 0 \Rightarrow \{\cot^{2} {(22 \frac{1}{2})}^{\circ} - 2 \cot {(22 \frac{1}{2})}^{\circ} + 1\} - 2 = 0 \Rightarrow {\{\cot {(22 \frac{1}{2})}^{\circ} - 1\}}^{2} = 2 \Rightarrow \cot {(22 \frac{1}{2})}^{\circ} - 1 = \sqrt{2} \Rightarrow \cot {(22 \frac{1}{2})}^{\circ} = \sqrt{2} + 1$

Page No 9.29:

Question 28:

(i) If $\cos x = - \frac{3}{5}$ and x lies in the IIIrd quadrant, find the values of
$\cos \frac{x}{2}, \sin \frac{x}{2}, \sin 2 x$ .

(ii) If $\cos x = - \frac{3}{5}$ and x lies in IInd quadrant, find the values of sin 2x and $\sin \frac{x}{2}$

Answer:

(i) $\cos x = - \frac{3}{5}$

Using the identity $\cos 2 θ = \cos^{2} θ - \sin^{2} θ$ , we get

$cosx = \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} \Rightarrow - \frac{3}{5} = 2 \cos^{2} \frac{x}{2} - 1 \Rightarrow 1 - \frac{3}{5} = 2 \cos^{2} \frac{x}{2} \Rightarrow \frac{2}{5} = 2 \cos^{2} \frac{x}{2} \Rightarrow \frac{1}{5} = \cos^{2} \frac{x}{2} \Rightarrow \cos \frac{x}{2} = \pm \sqrt{\frac{1}{5}}$

It is given that x lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$∴ \cos \frac{x}{2} = - \frac{1}{\sqrt{5}}$
Again,

$\cos x = \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} \Rightarrow - \frac{3}{5} = {(- \frac{1}{\sqrt{5}})}^{2} - \sin^{2} \frac{x}{2} \Rightarrow - \frac{3}{5} = \frac{1}{5} - \sin^{2} \frac{x}{2} \Rightarrow - \frac{1}{5} - \frac{3}{5} = - \sin^{2} \frac{x}{2} \Rightarrow \frac{4}{5} = \sin^{2} \frac{x}{2} \Rightarrow \sin \frac{x}{2} = \pm \frac{2}{\sqrt{5}}$

It is given that x lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$∴ \sin \frac{x}{2} = \frac{2}{\sqrt{5}}$

$Now, \sin x = \sqrt{1 - \cos^{2} x} \Rightarrow \sin x = {\sqrt{1 - (- \frac{3}{5})}}^{2} \sin x = \sqrt{1 - \frac{9}{25}} = \pm \frac{4}{5}$

Since x lies in the third quadrant, sinx is negative.

$∴ \sin x = - \frac{4}{5} \Rightarrow \sin 2 x = 2 \sin x \cos x \Rightarrow \sin 2 x = 2 \times (- \frac{4}{5}) \times (- \frac{3}{5}) \Rightarrow \sin 2 x = \frac{24}{25}$

(ii) $\cos x = - \frac{3}{5}$
$\sin x = \sqrt{1 - \cos^{2} x} = \sqrt{1 - (\frac{- 3}{5})} \Rightarrow \sin x = \pm \frac{4}{5}$
Here, x lies in the second quadrant.

$∴ \sin x = \frac{4}{5}$

We know,

sin2x = 2sinx cosx

$∴ \sin 2 x = 2 \times \frac{4}{5} \times (- \frac{3}{5}) = - \frac{24}{25}$
Now,
$\cos x = 1 - 2 \sin^{2} \frac{x}{2} \Rightarrow 2 \sin^{2} \frac{x}{2} = 1 - (- \frac{3}{5}) = \frac{8}{5} \Rightarrow \sin^{2} \frac{x}{2} = \frac{4}{5} \Rightarrow \sin \frac{x}{2} = \pm \frac{2}{\sqrt{5}}$

Since x lies in the second quadrant, $\frac{x}{2}$ lies in the first quadrant.
$∴ \sin \frac{x}{2} = \frac{2}{\sqrt{5}}$

Page No 9.29:

Question 29:

If $\sin x = \frac{\sqrt{5}}{3}$ and x lies in IInd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2} and \tan \frac{x}{2}$

Answer:

Given:
$\sin x = \frac{\sqrt{5}}{3}$
Using the identity $\cos x = \sqrt{1 - \sin^{2} x}$ , we get

$\cos x = \sqrt{1 - \sin^{2} x} = \sqrt{1 - {(\frac{\sqrt{5}}{3})}^{2}} = \pm \frac{2}{3}$

Since x lies in the 2nd quadrant, cosx is negative.

$∴ \cos x = - \frac{2}{3}$

Now,

$\cos x = 1 - 2 \sin^{2} \frac{x}{2} \Rightarrow - \frac{2}{3} = 1 - 2 \sin^{2} \frac{x}{2} \Rightarrow \sin \frac{x}{2} = \pm \sqrt{\frac{5}{6}}$

Since x lies in the 2nd quadrant, $\frac{x}{2}$ lies in the 1st quadrant.

$∴ \sin \frac{x}{2} = \sqrt{\frac{5}{6}}$

Again,

$\cos x = 2 \cos^{2} \frac{x}{2} - 1 \Rightarrow - \frac{2}{3} = 2 \cos^{2} \frac{x}{2} - 1 \Rightarrow \cos \frac{x}{2} = \pm \frac{1}{\sqrt{6}} \Rightarrow \cos \frac{x}{2} = \frac{1}{\sqrt{6}} (∵ \frac{x}{2} < \frac{π}{2})$
$Now, \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{\sqrt{5}}{\sqrt{6}}}{\frac{1}{\sqrt{6}}} = \sqrt{5}$

Page No 9.29:

Question 30:

(i) If 0 ≤ x ≤ π and x lies in the IInd quadrant such that $\sin x = \frac{1}{4}$ . Find the values of $\cos \frac{x}{2}, \sin \frac{x}{2} and \tan \frac{x}{2}$
(ii) If $\cos x = \frac{4}{5}$ and x is acute, find tan 2x
(iii) If $\sin x = \frac{4}{5}$ and $0 < x < \frac{π}{2}$ , find the value of sin 4x.

Answer:

(i) $\sin x = \frac{1}{4}$

$∴ \sin x = \sqrt{1 - \cos^{2} x} \Rightarrow {(\frac{1}{4})}^{2} = 1 - \cos^{2} x \Rightarrow \frac{1}{16} - 1 = - \cos^{2} x \Rightarrow \frac{15}{16} = \cos^{2} x \Rightarrow \cos x = \pm \frac{\sqrt{15}}{4}$

Since x lies in the 2nd quadrant, cos x is negative.

Thus,

$\cos x = - \frac{\sqrt{15}}{4}$

Now, using the identity $\cos x = 2 \cos^{2} \frac{x}{2} - 1$ , we get

$- \frac{\sqrt{15}}{4} = 2 \cos^{2} \frac{x}{2} - 1 \Rightarrow - \frac{\sqrt{15}}{8} = \cos^{2} \frac{x}{2} - \frac{1}{2} \Rightarrow \cos^{2} \frac{x}{2} = \frac{4 - \sqrt{15}}{8} \Rightarrow \cos \frac{x}{2} = \pm \frac{4 - \sqrt{15}}{8}$

Since x lies in the 2nd quadrant and $\frac{x}{2}$ lies in the 1st quadrant, $\cos \frac{x}{2}$ is positive.

$∴ \cos \frac{x}{2} = \frac{4 - \sqrt{15}}{8}$

Again,

$\cos x = \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} \Rightarrow - \frac{\sqrt{15}}{4} = {(\sqrt{\frac{4 - \sqrt{15}}{8}})}^{2} - \sin^{2} \frac{x}{2} \Rightarrow - \frac{\sqrt{15}}{4} = \frac{4 - \sqrt{15}}{8} - \sin^{2} \frac{x}{2} \Rightarrow \sin^{2} \frac{x}{2} = \frac{4 + \sqrt{15}}{8} \Rightarrow \sin \frac{x}{2} = \pm \sqrt{\frac{4 + \sqrt{15}}{8}} = \sqrt{\frac{4 + \sqrt{15}}{8}}$

Now,

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{\frac{4 + \sqrt{15}}{8}}}{\sqrt{\frac{4 - \sqrt{15}}{8}}} = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}} = \sqrt{\frac{(4 + \sqrt{15}) (4 + \sqrt{15})}{(4 - \sqrt{15}) (4 + \sqrt{15})}} = \frac{4 + \sqrt{15}}{4^{2} - {(\sqrt{15})}^{2}} = \frac{4 + \sqrt{15}}{16 - 15} = 4 + \sqrt{15}$

(ii) $\cos x = \frac{4}{5}$

$∴ \sin x = \sqrt{1 - \cos^{2} x} = \sqrt{1 - {(\frac{4}{5})}^{2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$

$∴ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

Now,
$\tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x} = \frac{2 (\frac{3}{4})}{1 - {(\frac{3}{4})}^{2}} = \frac{2 (\frac{3}{4})}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{24}{7}$

Hence, the value of tan 2x is $\frac{24}{7}$ .

(iii) $\sin x = \frac{4}{5}$ and $0 < x < \frac{π}{2}$ .

$∴ \sin x = \sqrt{1 - \cos^{2} x} \Rightarrow {(\frac{4}{5})}^{2} = 1 - \cos^{2} x \Rightarrow \frac{16}{25} - 1 = - \cos^{2} x \Rightarrow \frac{9}{25} = \cos^{2} x \Rightarrow \cos x = \pm \frac{3}{5}$

Since x lies in the 1st quadrant, cos x is positive.

Thus,

$\cos x = \frac{3}{5}$

Now,
$\sin (4 x) = 2 \sin (2 x) \cos (2 x) = 2 (2 \sin x \cos x) (1 - 2 \sin^{2} x) = 2 (2 \times \frac{4}{5} \times \frac{3}{5}) (1 - 2 {(\frac{4}{5})}^{2}) = 2 (\frac{24}{25}) (1 - \frac{32}{25}) = 2 (\frac{24}{25}) (\frac{25 - 32}{25}) = 2 (\frac{24}{25}) (\frac{- 7}{25}) = - \frac{336}{625}$

Hence, the value of sin 4x is $- \frac{336}{625}$ .

Page No 9.29:

Question 31:

If $\tan x = \frac{b}{a}$ , then find the value of $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}$ . [NCERT]

Answer:

Given: $\tan x = \frac{b}{a}$

$∴ \sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}} = \sqrt{\frac{1 + \tan x}{1 - \tan x}} + \sqrt{\frac{1 - \tan x}{1 + \tan x}} = \sqrt{\frac{1 + \frac{\sin x}{\cos x}}{1 - \frac{\sin x}{\cos x}}} + \sqrt{\frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}}}$
$= \sqrt{\frac{\cos x + \sin x}{\cos x - \sin x}} + \sqrt{\frac{\cos x - \sin x}{\cos x + \sin x}} = \frac{\cos x + \sin x + \cos x - \sin x}{\sqrt{(\cos x - \sin x) (\cos x + \sin x)}} = \frac{2 \cos x}{\sqrt{\cos^{2} x - \sin^{2} x}} = \frac{2 \cos x}{\sqrt{\cos 2 x}}$

Page No 9.29:

Question 32:

If $\tan A = \frac{1}{7}$ and $\tan B = \frac{1}{3}$ , show that cos 2A = sin 4B

Answer:

Given:
$\tan A = \frac{1}{7}$ and $\tan B = \frac{1}{3}$

Using the identity $\tan 2 B = \frac{2 \tan B}{1 - \tan^{2} B}$ , we get

$\tan 2 B = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} = \frac{3}{4}$
Now, using the identities $\cos 2 A = \frac{1 - \tan^{2} A}{1 + \tan^{2} A} and \sin 4 B = \frac{2 \tan 2 B}{1 + \tan^{2} 2 B}$ , we get

$\cos 2 A = \frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}} and \sin 4 B = \frac{2 \times \frac{3}{4}}{1 + {(\frac{3}{4})}^{2}} \Rightarrow \cos 2 A = \frac{48}{50} and \sin 4 B = \frac{2 \times \frac{3}{4} \times 16}{25} \Rightarrow \cos 2 A = \frac{24}{25} and \sin 4 B = \frac{24}{25}$

∴ cos 2A = sin 4B

Page No 9.29:

Question 33:

Prove that:
$\cos 7 ° \cos 14 ° \cos 28 ° \cos 56 ° = \frac{\sin 68 °}{16 \cos 83 °}$

Answer:

$LHS = \cos 7 ° \cos 14 ° \cos 28 ° \cos 56 °$

On dividing and multiplying by $2 \sin 7^{\circ}$ , we get

$= \frac{1}{2 \sin 7 °} \times 2 \sin 7 ° \times \cos 7 ° \times \cos 14 ° \times \cos 28 ° \times \cos 56 ° = \frac{2 \sin 14 °}{2 \times 2 \sin 7 °} \times \cos 14 ° \times \cos 28 ° \times \cos 56 ° = \frac{2 \sin 28 °}{2 \times 4 \sin 7 °} \times \cos 28 ° \times \cos 56 °$

$= \frac{2 \sin 56 °}{2 \times 8 \sin 7 °} \times \cos 56 ° = \frac{\sin 112 °}{16 \sin 7 °} = \frac{\sin (180 ° - 68 °)}{16 \sin (90 ° - 83 °)} = \frac{\sin 68 °}{16 \cos 83 °} [∵ \sin (180 ° - θ) = sinθ & \sin (90 ° - θ) = cosθ] = RHS Hence proved .$

Page No 9.29:

Question 34:

Prove that:
$\cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15} = \frac{1}{16}$

Answer:

$LHS = \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15}$

On dividing and multiplying by $2 \sin \frac{2 π}{15}$ , we get

$= \frac{1}{2 \sin \frac{2 π}{15}} \times (2 \sin \frac{2 π}{15} \times \cos \frac{2 π}{15}) \times \cos \frac{4 π}{15} \times \cos \frac{8 π}{15} \times \cos \frac{16 π}{15} = \frac{1}{2 \times 2 \sin \frac{2 π}{15}} \times (2 \sin \frac{4 π}{15} \times \cos \frac{4 π}{15}) \times \cos \frac{8 π}{15} \times \cos \frac{16 π}{15} = \frac{1}{2 \times 4 \sin \frac{2 π}{15}} (2 \sin \frac{8 π}{15} \times \cos \frac{8 π}{15}) \times \cos \frac{16 π}{15} = \frac{1}{2 \times 8 \sin \frac{2 π}{15}} (2 \sin \frac{16 π}{15} \times \cos \frac{16 π}{15}) = \frac{1}{16 \sin \frac{2 π}{15}} (\sin \frac{32 π}{15})$

$= - \frac{1}{16 \sin \frac{2 π}{15}} (\sin 2 π - \frac{32 π}{15}) [∵ \sin (2 π - θ) = - sinθ] = - \frac{1}{16 \sin \frac{2 π}{15}} \sin (- \frac{2 π}{15}) = \frac{1}{16} = RHS Hence proved .$

Page No 9.29:

Question 35:

Prove that: $\cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} = \frac{- 1}{16}$

Answer:

$\cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} = \frac{1}{2 \sin \frac{π}{5}} (2 \sin \frac{π}{5} \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}) (Multiplying and dividing by \frac{1}{2 \sin \frac{π}{5}}) = \frac{1}{2 \sin \frac{π}{5}} (\sin \frac{2 π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}) (\sin 2 A = 2 \sin A \cos A) = \frac{1}{4 \sin \frac{π}{5}} (2 \sin \frac{2 π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}) (Multiplying and dividing by 2)$
$= \frac{1}{4 \sin \frac{π}{5}} (\sin \frac{4 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}) = \frac{1}{8 \sin \frac{π}{5}} (2 \sin \frac{4 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}) (Multiplying and dividing by 2) = \frac{1}{8 \sin \frac{π}{5}} (\sin \frac{8 π}{5} \cos \frac{8 π}{5}) = \frac{1}{16 \sin \frac{π}{5}} (2 \sin \frac{8 π}{5} \cos \frac{8 π}{5}) (Multiplying and dividing by 2)$
$= \frac{\sin \frac{16 π}{5}}{16 \sin \frac{π}{5}} = \frac{\sin (3 π + \frac{π}{5})}{16 \sin \frac{π}{5}} = \frac{- \sin \frac{π}{5}}{16 \sin \frac{π}{5}} [\sin (3 π + θ) = - \sin θ] = \frac{- 1}{16}$

Page No 9.29:

Question 36:

Prove that:
$\cos \frac{π}{65} \cos \frac{2 π}{65} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65} = \frac{1}{64}$

Answer:

$LHS = \cos \frac{π}{65} \cos \frac{2 π}{65} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65}$

On dividing and multiplying by $2 \sin \frac{π}{65}$ , we get

$= \frac{1}{2 \sin \frac{π}{65}} \times 2 \sin \frac{π}{65} \times \cos \frac{π}{65} \times \cos \frac{2 π}{65} \times \cos \frac{4 π}{65} \times \cos \frac{8 π}{65} \times \cos \frac{16 π}{65} \times \cos \frac{32 π}{65} = \frac{2 \times \sin 2 \frac{π}{65}}{2 \times 2 \sin \frac{π}{65}} \times \cos \frac{2 π}{65} \times \cos \frac{4 π}{65} \times \cos \frac{8 π}{65} \times \cos \frac{16 π}{65} \times \cos \frac{32 π}{65} = \frac{2 \times \sin 4 \frac{π}{65}}{2 \times 4 \sin \frac{π}{65}} \times \cos \frac{4 π}{65} \times \cos \frac{8 π}{65} \times \cos \frac{16 π}{65} \times \cos \frac{32 π}{65} = \frac{2 \times \sin 8 \frac{π}{65}}{2 \times 8 \sin \frac{π}{65}} \times \cos \frac{8 π}{65} \times \cos \frac{16 π}{65} \times \cos \frac{32 π}{65}$

$= \frac{2 \times \sin 16 \frac{π}{65}}{2 \times 16 \sin \frac{π}{65}} \times \cos \frac{16 π}{65} \times \cos \frac{32 π}{65} = \frac{2 \times \sin 32 \frac{π}{65}}{2 \times 32 \sin \frac{π}{65}} \times \cos \frac{32 π}{65} = \frac{\sin 64 \frac{π}{65}}{64 \sin \frac{π}{65}} = \frac{\sin (π - \frac{π}{65})}{64 \sin \frac{π}{65}} = \frac{\sin \frac{π}{65}}{64 \sin \frac{π}{65}} [∵ \sin (π - θ) = sinθ] = \frac{1}{64} = RHS Hence proved .$

Page No 9.29:

Question 37:

If $2 \tan α = 3 \tan β,$ prove that $\tan (α - β) = \frac{\sin 2 β}{5 - \cos 2 β}$

Answer:

Given:
$2 \tan α = 3 \tan β$

$LHS = \frac{tanα - tanβ}{1 + tanα \times tanβ} = \frac{\frac{3}{2} \times tanβ - tanβ}{1 + \frac{3}{2} \tan^{2} β} (∵ 2 tanα = 3 tanβ) = \frac{\frac{1}{2} \times tanβ}{1 + \frac{3}{2} \tan^{2} β} = \frac{tanβ}{2 + 3 \tan^{2} β} = \frac{\frac{sinβ}{cosβ}}{2 + 3 \frac{\sin^{2} β}{\cos^{2} β}} = \frac{\frac{sinβ}{cosβ} \times \cos^{2} β}{2 \cos^{2} β + 3 \sin^{2} β}$

$= \frac{sinβ \times cosβ}{2 \cos^{2} β + 2 \sin^{2} β + \sin^{2} β} = \frac{1}{2} \frac{2 sinβ \times cosβ}{2 (\cos^{2} β + \sin^{2} β) + \sin^{2} β} = \frac{1}{2} \frac{\sin 2 β}{(2 + \sin^{2} β)} = \frac{\sin 2 β}{4 + 2 \sin^{2} β} = \frac{\sin 2 β}{4 + 2 (1 - \cos^{2} β)} = \frac{\sin 2 β}{6 - 2 \cos^{2} β} = \frac{\sin 2 β}{6 - (1 + \cos 2 β)} (∵ 1 + \cos 2 β = 2 \cos^{2} β) = \frac{\sin 2 β}{5 - \cos 2 β} = RHS Hence proved .$

Page No 9.29:

Question 38:

If $\sin α + \sin β = a and \cos α + \cos β = b$ , prove that

(i) $\sin (α + β) = \frac{2 a b}{a^{2} + b^{2}}$

(ii) $\cos (α - β) = \frac{a^{2} + b^{2} - 2}{2}$

Answer:

The given equations are $\sin α + \sin β = a and \cos α + \cos β = b$ .

(i)

$∵ \sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2} ∴ 2 \sin \frac{α + β}{2} \cos \frac{α - β}{2} = a . . . (1)$

Now, using the identity $\sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2}$ for the LHS of $\cos α + \cos β = b$ , we get

$2 \cos \frac{α + β}{2} \cos \frac{α - β}{2} = b . . . (2)$

On dividing (1) by (2), we get

$\tan \frac{α + β}{2} = \frac{a}{b}$

We know,

$sinθ = \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}$
$∴ \sin (α + β) = \frac{2 \tan (\frac{α + β}{2})}{1 + \tan^{2} (\frac{α + β}{2})} \Rightarrow \sin (α + β) = \frac{2 \times \frac{a}{b}}{1 + \frac{a^{2}}{b^{2}}} = \frac{2 a b}{a^{2} + b^{2}}$

(ii)

$On squaring sinα + sinβ = a and cosα + cosβ = b and adding them, we get \sin^{2} α + \sin^{2} β + 2 \times sinαsinβ + \cos^{2} α + \cos^{2} β + 2 \times cosαcosβ = a^{2} + b^{2} \Rightarrow 1 + 1 + 2 (sinαsinβ + cosαcosβ) = a^{2} + b^{2} \Rightarrow 2 (sinαsinβ + cosαcosβ) = a^{2} + b^{2} - 2 \Rightarrow 2 \cos (α - β) = a^{2} + b^{2} - 2 (∵ \cos (A - B) = sinAsinB + cosAcosB) \Rightarrow \cos (α - β) = \frac{a^{2} + b^{2} - 2}{2}$

Page No 9.29:

Question 39:

If $2 \tan \frac{α}{2} = \tan \frac{β}{2}$ , prove that $\cos α = \frac{3 + 5 \cos β}{5 + 3 \cos β}$

Answer:

$RHS = \frac{3 + 5 \cos β}{5 + 3 \cos β} = \frac{3 + 5 (\frac{1 - \tan^{2} \frac{β}{2}}{1 + \tan^{2} \frac{β}{2}})}{5 + 3 (\frac{1 - \tan^{2} \frac{β}{2}}{1 + \tan^{2} \frac{β}{2}})} = \frac{3 + 3 \tan^{2} \frac{β}{2} + 5 - 5 \tan^{2} \frac{β}{2}}{5 + 5 \tan^{2} \frac{β}{2} + 3 - 3 \tan \frac{β}{2}} = \frac{8 - 2 \tan^{2} \frac{β}{2}}{8 + 2 \tan^{2} \frac{β}{2}} = \frac{8 - 8 \tan^{2} \frac{α}{2}}{8 + 8 \tan^{2} \frac{α}{2}} [∵ 2 \tan \frac{α}{2} = \tan \frac{β}{2}] = \frac{8 (1 - \tan^{2} \frac{α}{2})}{8 (1 + \tan^{2} \frac{α}{2})} = \frac{1 - \tan^{2} \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} = \cos α = LHS Hence proved .$

Page No 9.29:

Question 40:

If $\cos x = \frac{\cos α + \cos β}{1 + \cos α \cos β}$ , prove that $\tan \frac{x}{2} = \pm \tan \frac{α}{2} \tan \frac{β}{2}$

Answer:

Given:

$\cos x = \frac{\cos α + \cos β}{1 + \cos α \cos β}$ ...(1)

$\Rightarrow \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} = \frac{cosα + cosβ}{1 + cosα \times cosβ} [∵ \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}] By componendo and dividendo, we get \frac{(1 - \tan^{2} \frac{x}{2}) + (1 + \tan^{2} \frac{x}{2})}{(1 - \tan^{2} \frac{x}{2}) - (1 + \tan^{2} \frac{x}{2})} = \frac{(1 + cosα \times cosβ + cosα + cosβ)}{- (1 + cosα cosβ - cosα - cosβ)} \Rightarrow \frac{2}{2 \tan^{2} \frac{x}{2}} = \frac{(1 + cosα) (1 + cosβ)}{(1 - cosα) (1 - cosβ)}$

$\Rightarrow \tan^{2} \frac{x}{2} = \frac{(1 - cosα) (1 - cosβ)}{(1 + cosα) (1 + cosβ)} \Rightarrow \tan^{2} \frac{x}{2} = \frac{2 \sin^{2} \frac{α}{2} \times 2 \sin^{2} \frac{β}{2}}{2 \cos^{2} \frac{α}{2} \times 2 \cos^{2} \frac{β}{2}} \Rightarrow \tan^{2} \frac{x}{2} = \tan^{2} \frac{α}{2} \times \tan^{2} \frac{β}{2} \Rightarrow \tan \frac{x}{2} = \pm \tan \frac{α}{2} \times \tan \frac{β}{2} Hence proved .$

Page No 9.29:

Question 41:

If $\sec (x + α) + \sec (x - α) = 2 \sec x$ , prove that $\cos x = \pm \sqrt{2} \cos \frac{α}{2}$

Answer:

Equation $\sec (x + α) + \sec (x - α) = 2 \sec x$ can be written as

$\frac{1}{\cos (x + α)} + \frac{1}{\cos (x - α)} = \frac{2}{\cos x} \Rightarrow \frac{1}{\cos x \times cosα - \sin x \times sinα} + \frac{1}{\cos x \times cosα + \sin x \times sinα} = \frac{2}{\cos x} [∵ \cos (A + B) = \cos A \times \cos B - \sin A \times \sin B and \cos (A - B) = \cos A \times \cos B + \sin A \times \sin B] \Rightarrow \frac{2 \cos x \times cosα}{\cos^{2} x \times \cos^{2} α - \sin^{2} x \times \sin^{2} α} = \frac{2}{\cos x} \Rightarrow \frac{\cos x \times cosα}{\cos^{2} x \times \cos^{2} α - (1 - \cos^{2} x) \times \sin^{2} α} = \frac{1}{\cos x}$

$\Rightarrow \frac{\cos^{2} x \times cosα}{\cos^{2} x \times \cos^{2} α - (1 - \cos^{2} x) \times \sin^{2} α} = 1 \Rightarrow \frac{\cos^{2} x \times cosα}{\cos^{2} x \times \cos^{2} α - \sin^{2} α + \cos^{2} x \sin^{2} α} = 1 \Rightarrow \cos^{2} x \times cosα = \cos^{2} x \times \cos^{2} α - \sin^{2} α + \cos^{2} x \sin^{2} α \Rightarrow \cos^{2} x \times cosα = \cos^{2} x (\cos^{2} α + \sin^{2} α) - \sin^{2} α \Rightarrow \cos^{2} x \times cosα = \cos^{2} x - \sin^{2} α$

$\Rightarrow \cos^{2} x \times cosα - \cos^{2} x = - \sin^{2} α \Rightarrow \cos^{2} x (cosα - 1) = - \sin^{2} α \Rightarrow \cos^{2} x (1 - cosα) = \sin^{2} α \Rightarrow \cos^{2} x = \frac{\sin^{2} α}{2 \sin^{2} \frac{α}{2}} (∵ 2 \sin^{2} \frac{x}{2} = 1 - \cos x)$

$\Rightarrow \cos^{2} x = \frac{4 \sin^{2} \frac{α}{2} \times \cos^{2} \frac{α}{2}}{2 \sin^{2} \frac{α}{2}} (∵ \sin^{2} x = 4 \sin^{2} \frac{x}{2} \times \cos^{2} \frac{x}{2}) \Rightarrow \cos x = \pm \sqrt{2} \cos \frac{α}{2} Hence proved .$

Page No 9.30:

Question 42:

If $\cos α + \cos β = \frac{1}{3}$ and sin $\sin α + \sin β = \frac{1}{4}$ , prove that $\cos \frac{α - β}{2} = \pm \frac{5}{24}$

Answer:

Squaring and adding equations $\cos α + \cos β = \frac{1}{3}$ and $\sin α + \sin β = \frac{1}{4}$ , we get

$(\cos^{2} α + \cos^{2} β + 2 cosα \times cosβ) + (\sin^{2} α + \sin^{2} β + 2 sinα \times sinβ) = \frac{1}{9} + \frac{1}{16} \Rightarrow 1 + 1 + 2 (cosα \times cosβ + sinα \times sinβ) = \frac{25}{144} \Rightarrow 2 + 2 \cos (α - β) = \frac{25}{144} (∵ \cos (A - B) = \cos A \times \cos B + \sin A \times \sin B) \Rightarrow \cos (α - β) = - \frac{263}{288} . . . (1)$

Now,

$\cos^{2} (\frac{α - β}{2}) = \frac{1 + \cos (α - β)}{2} = \frac{1 - \frac{263}{288}}{2} [From (1)] = \frac{25}{576} = \pm \frac{5}{24}$

Page No 9.30:

Question 43:

If $\sin α = \frac{4}{5} and \cos β = \frac{5}{13}$ , prove that $\cos \frac{α - β}{2} = \frac{8}{\sqrt{65}}$

Answer:

Given:
$\sin α = \frac{4}{5} \cos β = \frac{5}{13}$ .
Now,
$\cos α = \sqrt{1 - \sin^{2} α} = \sqrt{1 - {(\frac{4}{5})}^{2}} = \frac{3}{5} And, sinβ = \sqrt{1 - \cos^{2} α} = \sqrt{1 - {(\frac{5}{13})}^{2}} = \frac{12}{13}$

Now,

$\cos (α - β) = cosα \times cosβ + sinα \times sinβ \Rightarrow \cos (α - β) = \frac{3}{5} \times \frac{5}{13} \times \frac{4}{5} \times \frac{12}{13} = \frac{63}{65} Thus, \cos \frac{α - β}{2} = \sqrt{\frac{1 + \cos (α - β)}{2}} = \sqrt{\frac{1 + \frac{63}{65}}{2}} = \frac{8}{\sqrt{65}}$

Page No 9.30:

Question 44:

If $a \cos 2 x + b \sin 2 x = c$ has α and β as its roots, then prove that

(i) $\tan α + \tan β = \frac{2 b}{a + c}$ [NCERT EXEMPLAR]

(ii) $\tan α \tan β = \frac{c - a}{c + a}$

(iii) $\tan (α + β) = \frac{b}{a}$ [NCERT EXEMPLAR]

Answer:

Given: $a \cos 2 x + b \sin 2 x = c$

$\Rightarrow a (\frac{1 - \tan^{2} x}{1 + \tan^{2} x}) + b (\frac{2 \tan x}{1 + \tan^{2} x}) - c = 0 \Rightarrow a (1 - \tan^{2} x) + 2 b \tan x - c (1 + \tan^{2} x) = 0 \Rightarrow a - a \tan^{2} x + 2 b \tan x - c - c \tan^{2} x = 0 \Rightarrow (a + c) \tan^{2} x - 2 b \tan x + (c - a) = 0 . . . . . (1)$

This a quadratic equation in terms of $\tan^{2} x$ .

It is given that α and β are the roots of the given equation, so tanα and tanβ are the roots of (1).

Since tanα and tanβ are the roots of the equation $(a + c) \tan^{2} x - 2 b \tan x + (c - a) = 0$ . Therefore,

(i)
$\tan α + \tan β = - \frac{(- 2 b)}{a + c} (Sum of roots = - \frac{b}{a}) \Rightarrow \tan α + \tan β = \frac{2 b}{a + c}$

(ii)

$\tan α \tan β = \frac{c - a}{a + c} (Product of roots = \frac{c}{a}) Or \tan α \tan β = \frac{c - a}{c + a}$

(iii)
$\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} = \frac{\frac{2 b}{a + c}}{1 - (\frac{c - a}{c + a})} [From (i) and (ii)] = \frac{\frac{2 b}{a + c}}{\frac{c + a - c + a}{c + a}} = \frac{2 b}{2 a} = \frac{b}{a}$

Page No 9.30:

Question 45:

If $\cos α + \cos β = 0 = \sin α + \sin β$ , then prove that $\cos 2 α + \cos 2 β = - 2 \cos (α + β)$ . [NCERT EXEMPLAR]

Answer:

Given: $\cos α + \cos β = 0 = \sin α + \sin β$

$∴ \cos α + \cos β = 0$

Squaring on both sides, we get

$\cos^{2} α + \cos^{2} β + 2 \cos α \cos β = 0 . . . . . (1)$

Also,

$\sin α + \sin β = 0$

Squaring on both sides, we get

$\sin^{2} α + \sin^{2} β + 2 \sin α \sin β = 0 . . . . . (2)$

Subtracting (2) from (1), we get

$(\cos^{2} α + \cos^{2} β + 2 \cos α \cos β) - (\sin^{2} α + \sin^{2} β + 2 \sin α \sin β) = 0 \Rightarrow \cos^{2} α - \sin^{2} α + \cos^{2} β - \sin^{2} β + 2 (\cos α \cos β - \sin α \sin β) = 0 \Rightarrow \cos 2 α + \cos 2 β + 2 \cos (α + β) = 0 \Rightarrow \cos 2 α + \cos 2 β = - 2 \cos (α + β)$

Page No 9.36:

Question 1:

Prove that: $\sin 5 x = 5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x$

Answer:

$LHS = \sin 5 x = \sin (3 x + 2 x) = \sin 3 x \times \cos 2 x + \cos 3 x \times \sin 2 x = (3 \sin x - 4 \sin^{3} x) (1 - 2 \sin^{2} x) + (4 \cos^{3} x - 3 \cos x) \times 2 \sin x \cos x = 3 \sin x - 6 \sin^{3} x - 4 \sin^{3} x + 8 \sin^{5} x + (8 \cos^{4} x - 6 \cos^{2} x) \sin x = 3 \sin x - 10 \sin^{3} x + 8 \sin^{5} x + \{8 \sin x {(1 - \sin^{2} x)}^{2} - 6 \sin x (1 - \sin^{2} x)\} = 3 \sin x - 10 \sin^{3} x + 8 \sin^{5} x + \{8 \sin x (1 - 2 \sin^{2} x + \sin^{4} x) - 6 \sin x + 6 \sin^{3} x\} = 3 \sin x - 10 \sin^{3} x + 8 \sin^{5} x + 8 \sin x - 16 \sin^{3} x + 8 \sin^{5} x - 6 \sin x + 6 \sin^{3} x = 5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x = RHS Hence proved .$

Page No 9.36:

Question 2:

Prove that: $4 (\cos^{3} 10 ° + \sin^{3} 20 °) = 3 (\cos 10 ° + \sin 20 °)$

Answer:

$We know, \sin 60 ° = \cos 30 ° (= \frac{\sqrt{3}}{2}) \Rightarrow \sin 3 \times 20 ° = \cos 3 \times 10 ° \Rightarrow 3 \sin 20 ° - 4 \sin^{3} 20 ° = 4 \cos^{3} 10 ° - 3 \cos 10 ° (∵ \sin 3 θ = 3 sinθ - 4 \sin^{3} θ and \cos 3 θ = 4 \cos^{3} θ - 3 cosθ) \Rightarrow 4 (\cos^{3} 10 + \sin^{3} 20 °) = 3 (\cos 10 ° + \sin 20 °) Hence proved .$

Page No 9.36:

Question 3:

Prove that: $\cos^{3} x \sin 3 x + \sin^{3} x \cos 3 x = \frac{3}{4} \sin 4 x$

Answer:

$We know, \cos 3 x = 4 \cos^{3} x - 3 \cos x \Rightarrow \cos^{3} x = \frac{\cos 3 x + 3 \cos x}{4} . . . (i) Also, \sin 3 x = 3 \sin x - 4 \sin^{3} x \Rightarrow \sin^{3} x = \frac{3 \sin x - \sin 3 x}{4} . . . (ii)$
$Now, LHS = \cos^{3} x \sin 3 x + \sin^{3} x \cos 3 x = (\frac{\cos 3 x + 3 \cos x}{4}) \sin 3 x + (\frac{3 \sin x - \sin 3 x}{4}) \cos 3 x [Using (i) and (ii)] = \frac{1}{4} [3 (\sin 3 x \cos x + \sin x \cos 3 x) + (\cos 3 x \sin 3 x - \sin 3 x \cos 3 x)] = \frac{1}{4} [3 \sin (3 x + x) + 0] = \frac{3}{4} \sin 4 x = RHS Hence proved .$

Page No 9.36:

Question 4:

$\tan x \tan (x + \frac{π}{3}) + \tan x \tan (\frac{π}{3} - x) + \tan (x + \frac{π}{3}) \tan (x - \frac{π}{3}) = - 3$

Answer:

$\frac{π}{3} = 60 ° LHS = \tan x \tan (x + 60 °) + \tan x \tan (x - 60 °) + \tan (x + 60 °) \tan (x - 60 °) = \tan x (\frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x}) + \tan x (\frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x}) + (\frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x}) (\frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x}) [\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} and \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}] = \frac{(1 + \sqrt{3} \tan x) \tan x (\tan x + \sqrt{3}) + (1 - \sqrt{3} \tan x) \tan x (\tan x - \sqrt{3}) + \tan^{2} x - 3}{(1 + \sqrt{3} \tan x) (1 - \sqrt{3} \tan x)} = \frac{(\tan x + \sqrt{3} \tan^{2} x) (\tan x + \sqrt{3}) + (\tan x - \sqrt{3} \tan x) (\tan x - \sqrt{3}) + \tan^{2} x - 3}{1 - 3 \tan^{2} x}$

Page No 9.36:

Question 5:

$\tan x + \tan (\frac{π}{3} + x) - \tan (\frac{π}{3} - x) = 3 \tan 3 x$

Answer:

$\frac{π}{3} = 60 ° LHS = \tan x + \tan (60 ° + x) - \tan (60 ° - x) = \tan x + (\frac{\tan 60 ° + \tan x}{1 - \tan 60 ° \tan x}) - (\frac{\tan 60 ° - \tan x}{1 + \tan 60 ° \tan x}) [\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} and \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}] = \tan x + \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} - \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} = \tan x + \frac{\sqrt{3} + 3 \tan x + \tan x + \sqrt{3} \tan^{2} x + \sqrt{3} + 3 \tan x + \tan x - \sqrt{3} \tan^{2} x}{(1 - \sqrt{3} \tan x) (1 + \sqrt{3} \tan x)} = \tan x + \frac{8 \tan x}{1 - 3 \tan^{2} x}$

$= \frac{\tan x - 3 \tan^{3} x + 8 \tan x}{1 - 3 \tan^{2} x} = \frac{9 \tan x - 3 \tan^{3} x}{1 - 3 \tan^{2} x} = 3 (\frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x}) (∵ \tan 3 θ = \frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) = 3 \tan 3 x = RHS Hence proved .$

Page No 9.36:

Question 6:

$\cot x + \cot (\frac{π}{3} + x) + \cot (\frac{π}{3} - x) = 3 \cot 3 x$

Answer:

$\frac{π}{3} = 60 ° LHS = \cot x + \cot (60 ° + x) - \cot (60 ° - x) = \frac{1}{\tan x} + \frac{1}{\tan (60 ° + x)} - \frac{1}{\tan (60 ° - x)}$

$= \frac{1}{\tan x} + \frac{1 - \sqrt{3} \tan x}{\sqrt{3} + \tan x} - \frac{1 + \sqrt{3} \tan x}{\sqrt{3} - \tan x} [\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} and \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}] = \frac{1}{\tan x} - \frac{8 \tan x}{3 - \tan^{2} x} = \frac{3 - \tan^{2} x - 8 \tan^{2} x}{(\tan x) (3 - \tan^{2} x)} = \frac{3 - 9 \tan^{2} x}{3 \tan x - \tan^{3} x} = 3 (\frac{1 - 3 \tan^{2} x}{3 \tan x - \tan^{3} x}) = 3 \times \frac{1}{\tan 3 x} (∵ \tan 3 θ = \frac{1 - 3 \tan^{2} θ}{3 tanθ - \tan^{3} θ}) = 3 \cot 3 x = RHS Hence proved .$

Page No 9.36:

Question 7:

$\cot x + \cot (\frac{π}{3} + x) + \cot (\frac{2 π}{3} + x) = 3 \cot 3 x$

Answer:

$\frac{π}{3} = 60 °, \frac{2 π}{3} = 120 ° LHS = \cot x + \cot (60 ° + x) + \cot (120 ° + x) = \cot x + \cot (60 ° + x) - \cot [180 ° - (120 ° + x)] (∵ - cotθ = \cot (180 ° - θ)) = \cot x + \cot (60 ° + x) - \cot (60 ° - x) = \frac{1}{\tan x} + \frac{1}{\tan (60 ° + x)} - \frac{1}{\tan (60 ° - x)}$

$= \frac{1}{\tan x} + \frac{1 - \sqrt{3} \tan x}{\sqrt{3} + \tan x} - \frac{1 + \sqrt{3} \tan x}{\sqrt{3} - \tan x} [\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} and \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}] = \frac{1}{\tan x} - \frac{8 \tan x}{3 - \tan^{2} x} = \frac{3 - \tan^{2} x - 8 \tan^{2} x}{3 \tan x - \tan^{3} x} = \frac{3 - 9 \tan^{2} x}{3 \tan x - \tan^{3} x} = 3 (\frac{1 - 3 \tan^{2} x}{3 \tan x - \tan^{3} x}) = 3 \times \frac{1}{\tan 3 x} (∵ \tan 3 θ = \frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) = 3 \cot 3 x = RHS Hence proved .$

Page No 9.36:

Question 8:

$\sin 5 x = 5 \cos^{4} x \sin x - 10 \cos^{2} x \sin^{3} x + \sin^{5} x$

Answer:

$LHS = \sin 5 x = \sin (3 x + 2 x) = \sin 3 x \times \cos 2 x + \cos 3 x \times \sin 2 x = (3 \sin x - 4 \sin^{3} x) (2 \cos^{2} x - 1) + (4 \cos^{3} x - 3 \cos x) \times 2 \sin x \cos x = - 3 \sin x + 4 \sin^{3} x + 6 \sin x \cos^{2} x - 8 \sin^{3} x \cos^{2} x + 8 \sin x \cos^{4} x - 6 \sin x \cos^{2} x = 8 \sin x \cos^{4} x - 8 \sin^{3} x \cos^{2} x - 3 \sin x + 4 \sin^{3} x = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - 3 \sin x + 3 \sin x \cos^{4} x + 4 \sin^{3} x + 2 \sin^{3} x \cos^{2} x = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - 3 \sin x (1 - \cos^{4} x) + 2 \sin^{3} x (2 + \cos^{2} x)$

$= 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - 3 \sin x (1 - \cos^{2} x) (1 + \cos^{2} x) + 2 \sin^{3} x (2 + \cos^{2} x) = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - 3 \sin^{3} x (1 + \cos^{2} x) + 2 \sin^{3} x (2 + \cos^{2} x) = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - \sin^{3} x [3 (1 + \cos^{2} x) - 2 (2 + \cos^{2} x)] = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - \sin^{3} x [3 + 3 \cos^{2} x - 4 - 2 \cos^{2} x] = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - \sin^{3} x [\cos^{2} x - 1] = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x - \sin^{3} x \times (- \sin^{2} x) = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x = 5 \cos^{4} x \sin x - 10 \cos^{2} x \sin^{3} x + \sin^{5} x = RHS Hence proved .$

Page No 9.37:

Question 9:

$\sin^{3} x + \sin^{3} (\frac{2 π}{3} + x) + \sin^{3} (\frac{4 π}{3} + x) = - \frac{3}{4} \sin 3 x$

Answer:

$LHS = \sin^{3} x + \sin^{3} (\frac{2 π}{3} + x) + \sin^{3} (\frac{4 π}{3} + x) = \frac{3 \sin x - \sin 3 x}{4} + \frac{3 \sin (\frac{2 π}{3} + x) - \sin 3 (\frac{2 π}{3} + x)}{4} + \frac{3 \sin (\frac{4 π}{3} + x) - \sin 3 (\frac{4 π}{3} + x)}{4} [\sin^{3} θ = \frac{3 sinθ - \sin 3 θ}{4}] = \frac{3 \sin x - \sin 3 x}{4} + \frac{3 \sin \{π - (\frac{2 π}{3} + x)\} - \sin (2 π + 3 x)}{4} + \frac{3 \sin \{π + (\frac{π}{3} + x)\} - \sin (4 π + 3 x)}{4} = \frac{1}{4} [(3 \sin x - \sin 3 x) + \{3 \sin (\frac{π}{3} - x) - \sin 3 x\} - \{3 \sin (\frac{π}{3} + x) + \sin 3 x\}] = \frac{1}{4} [3 \sin x - \sin 3 x + 3 \sin (\frac{π}{3} - x) - 3 \sin (\frac{π}{3} + x) - \sin 3 x - \sin 3 x]$

$= \frac{1}{4} [3 \sin x - 3 \sin 3 x + 3 \{\sin (\frac{π}{3} - x) - \sin (\frac{π}{3} + x)\}] = \frac{1}{4} [3 \sin x - 3 \sin 3 x + 3 \{2 \cos \frac{\frac{π}{3} - x + \frac{π}{3} + x}{2} \sin \frac{\frac{π}{3} - x - \frac{π}{3} - x}{2}\}] [∵ sinC - sinD = 2 \cos \frac{C + D}{2} \sin \frac{C - D}{2}] = \frac{1}{4} [3 \sin x - 3 \sin 3 x + 6 \cos \frac{π}{3} \sin (- x)] = \frac{1}{4} [3 \sin x - 3 \sin 3 x - 3 \sin x] = - \frac{3}{4} \sin x = RHS Hence proved .$

Page No 9.37:

Question 10:

Prove that $|\sin x \sin (\frac{π}{3} - x) \sin (\frac{π}{3} + x)| \leq \frac{1}{4}$ for all values of x

Answer:

$\frac{π}{3} = 60 ° We have, |\sin x \sin (60 - x) \sin (60 + x)| = |\sin x (\sin^{2} 60 - \sin^{2} x)| [∵ \sin (A + B) \sin (A - B) = \sin^{2} A - \sin^{2} B] = |\sin x (\frac{3}{4} - \sin^{2} x)| = |\frac{1}{4} \sin x (3 - 4 \sin^{2} x)| = |\frac{1}{4} 3 \sin x - 4 \sin^{3} x| = \frac{1}{4} |\sin 3 x| (∵ 3 \sin x - 4 \sin^{3} x = \sin 3 x) \leq \frac{1}{4} (∵ |\sin x| \leq 1 for all x) ∴ |\sin x \sin (60 - x) \sin (60 + x)| \leq \frac{1}{4} Hence proved .$

Page No 9.37:

Question 11:

Prove that $|\cos x \cos (\frac{π}{3} - x) \cos (\frac{π}{3} + x)| \leq \frac{1}{4}$ for all values of x

Answer:

$\frac{π}{3} = 60 ° We have, |\cos x \cos (60 ° - x) \cos (60 ° + x)| = |\cos x (\cos^{2} 60 ° - \sin^{2} x)| [\cos^{2} A - \sin^{2} B = \cos (A - B) \cos (A + B)] = |\cos x (\frac{1}{4} - \sin^{2} x)| = |\cos x \frac{1}{4} (1 - 4 \sin^{2} x)| = |\frac{1}{4} \cos x \{1 - 4 (1 - \cos^{2} x)\}| = |\frac{1}{4} \cos x \{- 3 + 4 \cos^{2} x\}|$

$= |\frac{1}{4} (4 \cos^{3} x - 3 \cos x)| = |\frac{1}{4} \cos 3 x| [∵ \cos 3 x = 4 \cos^{3} x - 3 \cos x] \leq \frac{1}{4} (∵ |\cos x| \leq 1 for all x) ∴ |\cos x \cos (60 ° - x) \cos (60 ° + x)| \leq \frac{1}{4}$

Page No 9.42:

Question 1:

Prove that:
$\sin^{2} \frac{2 π}{5} - \sin^{2 -} \frac{π}{3} = \frac{\sqrt{5} - 1}{8}$

Answer:

$\frac{2 π}{5} = 72 °, \frac{π}{3} = 60 ° LHS = \sin^{2} 72 ° - \sin^{2} 60 ° = \sin^{2} (90 ° - 18 °) - \frac{3}{4} = \cos^{2} 18 ° - \frac{3}{4} (∵ \sin (90 ° - θ) = cosθ) = {(\frac{\sqrt{10 + 2 \sqrt{5}}}{4})}^{2} - \frac{3}{4} (∵ \cos 18 ° = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}) = \frac{10 + 2 \sqrt{5}}{16} - \frac{3}{4}$

$= \frac{10 + 2 \sqrt{5} - 12}{16} = \frac{2 \sqrt{5} - 2}{16} = \frac{\sqrt{5} - 1}{8} = RHS Hence proved .$

Page No 9.42:

Question 2:

Prove that:
$\sin^{2} 24 ° - \sin^{2} 6 ° = \frac{\sqrt{5} - 1}{8}$

Answer:

$LHS = \sin^{2} 24 ° - \sin^{2} 6 ° = \sin (24 ° + 6 °) \sin (24 ° - 6 °) [\sin (A + B) \sin (A - B) = \sin^{2} A - \sin^{2} B] = \sin 30 ° \sin 18 ° = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} (∵ \sin 18 ° = \frac{\sqrt{5} - 1}{4}) = \frac{\sqrt{5} - 1}{8} = RHS Hence proved .$

Page No 9.42:

Question 3:

Prove that:
$\sin^{2} 42 ° - \cos^{2} 78 = \frac{\sqrt{5} + 1}{8}$

Answer:

$LHS = \sin^{2} 42 ° - \cos^{2} 78 ° = \sin^{2} (90 ° - 48 °) - \cos^{2} (90 ° - 12 °) = \cos^{2} 48 ° - \sin^{2} 12 ° = \cos (48 ° + 12 °) \cos (48 ° - 12 °) [\cos (A + B) \cos (A - B) = \cos^{2} A - \sin^{2}] = \cos 60 ° \cos 36 ° = \frac{1}{2} \times \frac{\sqrt{5} + 1}{4} (∵ \cos 36 ° = \frac{\sqrt{5} + 1}{4}) = \frac{\sqrt{5} + 1}{8} = RHS Hence proved .$

Page No 9.42:

Question 4:

Prove that:
$\cos 78 ° \cos 42 ° \cos 36 ° = \frac{1}{8}$

Answer:

$LHS = \cos 78 ° \cos 42 ° \cos 36 ° = \frac{(2 \cos 78 ° \cos 42 °)}{2} \cos 36 ° = \frac{\cos (78 ° + 42 °) + \cos (78 ° - 42 °)}{2} \times \cos 36 ° [2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \frac{1}{2} (\cos 120 ° + \cos 36 °) \cos 36 °$

$= \frac{1}{2} (- \cos (180 ° - 120 °) + \cos 36 °) \cos 36 ° = \frac{1}{2} (- \cos 60 ° + \cos 36 °) \cos 36 ° = \frac{1}{2} (- \frac{1}{2} + \frac{\sqrt{5} + 1}{4}) \frac{\sqrt{5} + 1}{4} = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{5} + 1}{4} = \frac{1}{8} = RHS Hence proved .$

Page No 9.42:

Question 5:

Prove that:
$\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15} = \frac{1}{16}$

Answer:

$LHS = \cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15}$

     $= \frac{2 \sin \frac{π}{15} \cos \frac{π}{15}}{2 \sin \frac{π}{15}} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15} [On dividing and multiplying by 2 \sin \frac{π}{15}]$

    $= \frac{2 \sin \frac{2 π}{15} \times \cos \frac{2 π}{15}}{2 \times 2 \sin \frac{π}{15}} \cos \frac{4 π}{15} \cos \frac{7 π}{15} = \frac{2 \sin \frac{4 π}{15} \times \cos \frac{4 π}{15}}{2 \times 2 \times 2 \sin \frac{π}{15}} \cos \frac{7 π}{15} = \frac{\sin \frac{8 π}{15}}{2 \times 2 \times 2 \sin \frac{π}{15}} \cos \frac{7 π}{15}$

   $= \frac{2 \sin \frac{8 π}{15} \cos \frac{7 π}{15}}{2 \times 2 \times 2 \times 2 \sin \frac{π}{15}} = \frac{2 \sin \frac{8 π}{15} \cos \frac{7 π}{15}}{16 \sin \frac{π}{15}} = \frac{\sin (\frac{8 π}{15} + \frac{7 π}{15}) + \sin (\frac{8 π}{15} - \frac{7 π}{15})}{16 \sin \frac{π}{15}} [∵ 2 \sin A \cos B = \sin (A + B) + \sin (A - B)]$

   $= \frac{sinπ + \sin \frac{π}{15}}{16 \sin \frac{π}{15}} = \frac{0 + \sin \frac{π}{15}}{16 \sin \frac{π}{15}} = \frac{\sin \frac{π}{15}}{16 \sin \frac{π}{15}} = \frac{1}{16} = RHS$
Hence proved.

Page No 9.42:

Question 6:

Prove that:
$\cos 6 ° \cos 42 ° \cos 66 ° \cos 78 ° = \frac{1}{16}$

Answer:

$LHS = \cos 6 ° \cos 42 ° \cos 66 ° \cos 78 ° = \frac{1}{4} (2 \cos 6 ° \cos 66 °) (2 \cos 42 ° \cos 78 °) = \frac{1}{4} (\cos 72 ° + \cos 60 °) (\cos 120 ° + \cos 36 °) [∵ 2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \frac{1}{4} \{\cos (90 ° - 72 °) + \frac{1}{2}\} \{- \frac{1}{2} + \frac{\sqrt{5} + 1}{4}\}$
$= \frac{1}{4} (\sin 18 ° + \frac{1}{2}) (- \frac{1}{2} + \frac{\sqrt{5} + 1}{4}) = \frac{1}{4} (\frac{\sqrt{5} - 1}{4} + \frac{1}{2}) (\frac{\sqrt{5} + 1}{4} - \frac{1}{2}) = \frac{1}{4} (\frac{\sqrt{5} - 1 + 2}{4}) (\frac{\sqrt{5} + 1 - 2}{4}) = \frac{1}{64} (\sqrt{5} + 1) (\sqrt{5} - 1) = \frac{1}{64} (5 - 1) = \frac{1}{16} = RHS Hence proved .$

Page No 9.42:

Question 7:

Prove that:
$\sin 6 ° \sin 42 ° \sin 66 ° \sin 78 ° = \frac{1}{16}$

Answer:

$LHS = \frac{1}{4} (2 \sin 6 ° \sin 66 °) (2 \sin 42 ° \sin 78 °) = \frac{1}{4} (\cos 60 ° - \cos 72 °) (\cos 36 ° - \cos 120 °) [2 \sin A \sin B = \cos (A - B) - \cos (A + B)] = \frac{1}{4} (\frac{1}{2} - \sin 18 °) (\frac{\sqrt{5} + 1}{4} + \frac{1}{2})$

$= \frac{1}{4} (\frac{1}{2} - \frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4} + \frac{1}{2}) = \frac{1}{4} (\frac{2 - \sqrt{5} + 1}{4}) (\frac{\sqrt{5} + 1 + 2}{4}) = \frac{1}{64} (3 - \sqrt{5}) (3 + \sqrt{5}) = \frac{1}{64} (9 - 5) = \frac{1}{16} = RHS Hence proved .$

Page No 9.42:

Question 8:

Prove that:
$\cos 36 ° \cos 42 ° \cos 60 ° \cos 78 ° = \frac{1}{16}$

Answer:

$LHS = \cos 36 ° \cos 42 ° \cos 60 ° \cos 78 ° = \frac{1}{2} \cos 36 ° \cos 60 ° (2 \cos 42 ° \cos 78 °) [2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \frac{1}{2} (\frac{\sqrt{5} + 1}{4}) \times \frac{1}{2} (\cos 120 ° + \cos 36 °) = (\frac{\sqrt{5} + 1}{16}) (- \frac{1}{2} + \frac{\sqrt{5} + 1}{4}) = \frac{(\sqrt{5} + 1) (\sqrt{5} - 1)}{64} = \frac{5 - 1}{64} = \frac{1}{16} = RHS Hence proved .$

Page No 9.42:

Question 9:

Prove that :
$\sin \frac{π}{5} \sin \frac{2 π}{5} \sin \frac{3 π}{5} \sin \frac{4 π}{5} = \frac{5}{16}$

Answer:

$LHS = \sin \frac{π}{5} \sin \frac{2 π}{5} \sin \frac{3 π}{5} \sin \frac{4 π}{5} = \frac{1}{2} (2 \sin \frac{π}{5} \sin \frac{4 π}{5}) \frac{1}{2} (2 \sin \frac{2 π}{5} \sin \frac{3 π}{5}) = \frac{1}{4} (\cos (\frac{π}{5} - \frac{4 π}{5}) - \cos (\frac{π}{5} + \frac{4 π}{5})) (\cos (\frac{2 π}{5} - \frac{3 π}{5}) - \cos (\frac{2 π}{5} + \frac{3 π}{5})) = \frac{1}{4} (\cos (\frac{- 3 π}{5}) - \cos (\frac{5 π}{5})) (\cos (\frac{- π}{5}) - \cos (\frac{5 π}{5})) = \frac{1}{4} (\cos (\frac{3 π}{5}) - \cos (π)) (\cos (\frac{π}{5}) - \cos (π)) = \frac{1}{4} (\cos (\frac{3 π}{5}) + 1) (\cos (\frac{π}{5}) + 1) = \frac{1}{4} (\cos (π - \frac{2 π}{5}) + 1) (\cos (\frac{π}{5}) + 1) = \frac{1}{4} (- \cos (\frac{2 π}{5}) + 1) ((\frac{\sqrt{5} + 1}{4}) + 1) (∵ \cos \frac{π}{5} = \frac{\sqrt{5} + 1}{4}) = \frac{1}{4} (- (\frac{\sqrt{5} - 1}{4}) + 1) ((\frac{\sqrt{5} + 1}{4}) + 1) (∵ \cos \frac{2 π}{5} = \frac{\sqrt{5} - 1}{4}) = \frac{1}{4} (- (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4}) - (\frac{\sqrt{5} - 1}{4}) + (\frac{\sqrt{5} + 1}{4}) + 1) = \frac{1}{4} (- (\frac{{(\sqrt{5})}^{2} - 1}{16}) + (\frac{\sqrt{5} + 1 - \sqrt{5} + 1}{4}) + 1) = \frac{1}{4} (- (\frac{4}{16}) + (\frac{2}{4}) + 1) = \frac{1}{4} (- \frac{1}{4} + \frac{2}{4} + 1) = \frac{1}{4} (\frac{- 1 + 2 + 4}{4}) = \frac{5}{16} = RHS$

Thus, LHS = RHS

Hence, $\sin \frac{π}{5} \sin \frac{2 π}{5} \sin \frac{3 π}{5} \sin \frac{4 π}{5} = \frac{5}{16}$ .

Page No 9.42:

Question 10:

Prove that:
$\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{3 π}{15} \cos \frac{4 π}{15} \cos \frac{5 π}{15} \cos \frac{6 π}{15} \cos \frac{7 π}{15} = \frac{1}{128}$

Answer:

$LHS = \cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{3 π}{15} \cos \frac{5 π}{15} \cos \frac{6 π}{15} \cos \frac{7 π}{15} = \cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} (\cos \frac{3 π}{15} \cos \frac{6 π}{15}) \times (- \cos \frac{8 π}{15}) = - \frac{1}{2} [\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15}] \times \frac{1}{2} \times (\cos \frac{3 π}{15} \cos \frac{6 π}{15}) = - \frac{1}{2} \times \frac{2^{3}}{2^{4} \sin \frac{π}{15}} [2 \sin \frac{π}{15} \cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15}] \times \frac{2}{2^{2} \times \sin \frac{3 π}{15}} (2 \sin \frac{3 π}{15} \cos \frac{3 π}{15} \cos \frac{6 π}{15}) = - \frac{2^{3}}{132 \sin \frac{π}{15}} [\sin \frac{2 π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15}] \times \frac{2}{4 \sin \frac{3 π}{15}} (\sin \frac{6 π}{15} \cos \frac{6 π}{15}) = - \frac{2^{2}}{32 \sin \frac{π}{15}} [2 \sin \frac{2 π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15}] \times \frac{1}{4 \sin \frac{3 π}{15}} (2 \sin \frac{6 π}{15} \cos \frac{6 π}{15})$
$= - \frac{2}{32 \sin \frac{π}{15}} [\sin \frac{8 π}{15} \cos \frac{8 π}{15}] \times \frac{\sin \frac{12 π}{15}}{4 \sin \frac{3 π}{15}} = - \frac{1}{32 \sin \frac{π}{15}} [\sin \frac{16 π}{15}] \times \frac{\sin \frac{12 π}{15}}{4 \sin \frac{3 π}{15}} = - \frac{\sin (π + \frac{π}{15})}{128 \sin \frac{π}{15}} \times \frac{\sin (π - \frac{3 π}{15})}{\sin \frac{3 π}{15}} = - \frac{- \sin \frac{π}{15}}{128 \sin \frac{π}{15}} \times \frac{\sin \frac{3 π}{15}}{\sin \frac{3 π}{15}} = \frac{1}{128} = RHS Hence proved .$

Page No 9.42:

Question 1:

If $\cos 4 x = 1 + k \sin^{2} x \cos^{2} x$ , then write the value of k.

Answer:

$We have, \cos 4 x = 1 + k \sin^{2} x \cos^{2} x \Rightarrow \cos (2 \times 2 x) = 1 + k \sin^{2} x \cos^{2} x \Rightarrow 1 - 2 \sin^{2} 2 x = 1 + k \sin^{2} x \cos^{2} x \Rightarrow 1 - 2 {(2 sinxcos x)}^{2} = 1 + k \sin^{2} x \cos^{2} x \Rightarrow 1 - 8 \sin^{2} x \cos^{2} x = 1 + k \sin^{2} x \cos^{2} x \Rightarrow \sin^{2} {xcos}^{2} x (k + 8) = 0 \Rightarrow k + 8 = 0 ∴ k = - 8$

Page No 9.42:

Question 2:

If $\tan \frac{x}{2} = \frac{m}{n}$ , then write the value of m sin x + n cos x.

Answer:

Given:

$\tan \frac{x}{2} = \frac{m}{n}$
$\Rightarrow \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{m}{n} Let \sin \frac{x}{2} be m k and \cos \frac{x}{2} be n k . Now, m \sin x + n \cos x = 2 m \sin \frac{x}{2} \cos \frac{x}{2} + n (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}) = 2 m \times m k \times n k + n (n^{2} k^{2} - m^{2} k^{2})$

$= 2 m^{2} k^{2} n + n k^{2} (n^{2} - m^{2}) = n k^{2} (2 m^{2} + n^{2} - m^{2}) = n k^{2} (m^{2} + n^{2}) = n (m^{2} k^{2} + n^{2} k^{2}) = n (\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2}) = n (1) ∴ m \sin x + n \cos x = n$

Page No 9.42:

Question 3:

If $\frac{π}{2} < x < \frac{3 π}{2}$ , then write the value of $\sqrt{\frac{1 + \cos 2 x}{2}}$ .

Answer:

$∵ \frac{π}{2} < x < \frac{3 π}{2} ∴ \sqrt{\frac{1 + \cos 2 x}{2}} = \sqrt{\frac{2 \cos^{2} x}{2}} = |\cos x| In second quadrant \cos x is negative . ∴ \sqrt{\frac{1 + \cos 2 x}{2}} = - \cos x$

Page No 9.42:

Question 4:

If $\frac{π}{2} < x < π,$ the write the value of $\sqrt{2 + \sqrt{2 + 2 \cos 2 x}}$ in the simplest form.

Answer:

We have,

$\sqrt{2 + \sqrt{2 + 2 \cos 2 x}} = \sqrt{2 + \sqrt{2 (1 + \cos 2 x)}} = \sqrt{2 + \sqrt{2.2 \cos^{2} x}} = \sqrt{2 + 2 |\cos x|} = \sqrt{2 - 2 \cos x} (∵ \frac{π}{2} < x < π) = \sqrt{2 (1 - \cos x)} = \sqrt{2.2 \sin^{2} \frac{x}{2}} = 2 |\sin \frac{x}{2}| = 2 \sin \frac{x}{2} (∵ \frac{π}{4} < \frac{x}{2} < \frac{π}{2})$

Page No 9.42:

Question 5:

If $\frac{π}{2} < x < π$ , then write the value of $\frac{\sqrt{1 - \cos 2 x}}{1 + \cos 2 x}$ .

Answer:

We have,
$\begin{array}{rcl} \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} & = & \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x}} \\ = & \frac{|\sin x|}{|\cos x|} \\ = & \frac{\sin x}{- \cos x} (∵ \frac{π}{2} < x < π) \\ = & - \tan x \end{array}$

Page No 9.42:

Question 6:

If $π < x < \frac{3 π}{2}$ , then write the value of $\sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}}$ .

Answer:

$We have, \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} = \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x}} = \frac{|\sin x|}{|\cos x|} = \frac{|\sin x|}{|\cos x|} = \frac{- \sin x}{- \cos x} (∵ π < x < \frac{3 π}{2}) ∴ \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} = \tan x$

Page No 9.42:

Question 7:

In a right angled triangle ABC, write the value of sin² A + Sin² B + Sin² C.

Answer:

$Let, ∠ B = 90 ° ∴ A + C = 90 ° = \frac{π}{2} \Rightarrow C = \frac{π}{2} - A \Rightarrow \sin C = \sin (\frac{π}{2} - A) \Rightarrow \sin C = \cos A . . . (i) Now, \sin^{2} A + \sin^{2} B + \sin^{2} C = \sin^{2} A + 1 + \sin^{2} C (∵ \sin B = \sin 90 ° = 1) = \sin^{2} A + \cos^{2} A + 1 [Using (i)] = 1 + 1 = 2$

Page No 9.42:

Question 8:

Write the value of $\cos^{2} 76 ° + \cos^{2} 16 ° - \cos 76 ° \cos 16 °$ .

Answer:

$We have, \cos^{2} 76 ° + \cos^{2} 16 ° - \cos 76 ° \cos 16 ° = \frac{1}{2} [1 + \cos 2 (76) ° + 1 + \cos 2 (16) ° - \cos (76 + 16) ° - \cos (76 - 16) °] [∵ 2 \cos^{2} θ = 1 + \cos 2 θ and 2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \frac{1}{2} [2 + \cos 152 ° + \cos 32 ° - \cos 92 ° - \frac{1}{2}] = \frac{1}{2} [\frac{3}{2} - \cos (180 - 152 °) + \cos 32 ° - \cos 92 °]$

$= \frac{1}{2} [\frac{3}{2} - \cos 28 ° + 2 \sin \frac{92 ° + 32 °}{2} \sin \frac{92 ° - 32 °}{2}] [\cos C - \cos D = 2 \sin \frac{C + D}{2} \sin \frac{D - C}{2}] = \frac{1}{2} [\frac{3}{2} - \cos 28 ° + 2 \sin \frac{124 °}{2} \sin \frac{60 °}{2}] = \frac{1}{2} [\frac{3}{2} - \cos 28 ° + 2 \sin 62 ° \sin 30 °] = \frac{1}{2} [\frac{3}{2} - \cos 28 ° + 2 \sin 62 ° \times \frac{1}{2}] = \frac{1}{2} [\frac{3}{2} - \cos 28 ° + \sin 62 °] = \frac{1}{2} [\frac{3}{2} - \cos 28 ° + \sin (90 - 28) °] = \frac{1}{2} [\frac{3}{2} - \cos 28 ° + \cos 28 °] = \frac{3}{4}$

Page No 9.42:

Question 9:

If $\frac{π}{4} < x < \frac{π}{2}$ , then write the value of $\sqrt{1 - \sin 2 x}$ .

Answer:

$We have, \sqrt{1 - \sin 2 x} = \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x} = \sqrt{{(\sin x - \cos x)}^{2}} = |\sin x - \cos x| = \sin x - \cos x [∵ \sin x > \cos x for \frac{π}{4} < x < \frac{π}{2}] ∴ \sqrt{1 - \sin 2 x} = \sin x - \cos x$

Page No 9.42:

Question 10:

Write the value of $\cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} .$

Answer:

$We have, \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} = \frac{2 \sin \frac{π}{7} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{π}{7}} [On dividing and multiplying by 2 \sin \frac{π}{7}] = \frac{2 \times \sin \frac{2 π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7}}{2 \times 2 \sin \frac{π}{7}}$

Proceeding in the same way, we get

$\cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} = \frac{\sin \frac{8 π}{7}}{8 \sin \frac{π}{7}} = \frac{\sin (π + \frac{π}{7})}{8 \sin \frac{π}{7}} = \frac{- \sin \frac{π}{7}}{8 \sin \frac{π}{7}} ∴ \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} = \frac{- 1}{8}$

Page No 9.42:

Question 11:

If $\tan A = \frac{1 - \cos B}{\sin B}$ , then find the value of tan2A.

Answer:

Given:

$\tan A = \frac{1 - \cos B}{\sin B} \Rightarrow \tan A = \frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}} (1 - \cos 2 θ = 2 \sin^{2} θ and \sin 2 θ = 2 \sin θ \cos θ) \Rightarrow \tan A = \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} = \tan \frac{B}{2} \Rightarrow A = \frac{B}{2}$

$\Rightarrow 2 A = B ∴ \tan 2 A = \tan B$

Hence, the value of tan2A is tanB.

Page No 9.42:

Question 12:

If $\sin x + \cos x = a$ , then find the value of $\sin^{6} x + \cos^{6} x$ .

Answer:

Given: $\sin x + \cos x = a$

Squaring on both sides, we get

$\sin^{2} x + \cos^{2} x + 2 \sin x \cos x = a^{2} \Rightarrow 1 + 2 \sin x \cos x = a^{2} \Rightarrow \sin x \cos x = \frac{a^{2} - 1}{2} . . . . . (1)$

Now,

$\sin^{6} x + \cos^{6} x = {(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x) = 1 - 3 {(\frac{a^{2} - 1}{2})}^{2} [Using (1)] = \frac{4 - 3 {(a^{2} - 1)}^{2}}{4}$

Hence, the required value is $\frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}]$ .

Page No 9.42:

Question 13:

If $\sin x + \cos x = a$ , find the value of $|\sin x - \cos x|$ .

Answer:

Given: $\sin x + \cos x = a$

Now,

${(\sin x + \cos x)}^{2} + {(\sin x - \cos x)}^{2} = \sin^{2} x + \cos^{2} x + 2 \sin x \cos x + \sin^{2} x + \cos^{2} x - 2 \sin x \cos x \Rightarrow {(\sin x + \cos x)}^{2} + {(\sin x - \cos x)}^{2} = 2 (\sin^{2} x + \cos^{2} x) \Rightarrow {(\sin x + \cos x)}^{2} + {(\sin x - \cos x)}^{2} = 2$

$∴ a^{2} + {(\sin x - \cos x)}^{2} = 2 \Rightarrow {(\sin x - \cos x)}^{2} = 2 - a^{2} \Rightarrow \sqrt{{(\sin x - \cos x)}^{2}} = \sqrt{2 - a^{2}} \Rightarrow |\sin x - \cos x| = \sqrt{2 - a^{2}} (\sqrt{x^{2}} = |x|)$

Thus, the required value is $\sqrt{2 - a^{2}}$ .

Page No 9.43:

Question 1:

$8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$ is equal to

(a) 8 cos x
(b) cos x
(c) 8 sin x
(d) sin x

Answer:

(d) sin x

$We have, 8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} = 4 \times (2 \sin \frac{x}{8} \cos \frac{x}{8}) \cos \frac{x}{2} \cos \frac{x}{4} = 4 \times \sin \frac{x}{4} \cos \frac{x}{2} \cos \frac{x}{4} = 2 \times (2 \sin \frac{x}{4} \cos \frac{x}{4}) \cos \frac{x}{2} = 2 \times \sin \frac{x}{2} \cos \frac{x}{2} = \sin x$

Page No 9.43:

Question 2:

$\frac{\sec 8 A - 1}{\sec 4 A - 1} =$

(a) $\frac{\tan 2 A}{\tan 8 A}$

(b) $\frac{\tan 8 A}{\tan 2 A}$

(c) $\frac{\cot 8 A}{\cot 2 A}$

(d) none of these.

Answer:

(b) $\frac{\tan 8 A}{\tan 2 A}$

$We have, \frac{\sec 8 A - 1}{\sec 4 A - 1} = \frac{\frac{1}{\cos 8 A} - 1}{\frac{1}{\cos 4 A} - 1} = \frac{\cos 4 A}{\cos 8 A} \times \frac{1 - \cos 8 A}{1 - \cos 4 A}$

$= \frac{\cos 4 A}{\cos 8 A} \times \frac{2 \sin^{2} 4 A}{2 \sin^{2} 2 A} (2 \sin^{2} θ = 1 - \cos 2 θ) = \frac{(2 \cos 4 A \sin 4 A) \sin 4 A}{2 \times \cos 8 A \sin^{2} 2 A} = \frac{\sin 8 A \sin 4 A}{\cos 8 A \times 2 \sin 2 A \times \sin 2 A} = \tan 8 A \times \frac{2 \sin 2 A \times \cos 2 A}{2 \sin 2 A \times \sin 2 A} = \tan 8 A \times \cot 2 A = \frac{\tan 8 A}{\tan 2 A}$

Page No 9.43:

Question 3:

The value of $\cos \frac{π}{65} \cos \frac{2 π}{65} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65}$ is
(a) $\frac{1}{8}$

(b) $\frac{1}{16}$

(c) $\frac{1}{32}$

(d) none of these

Answer:

(d) none of these

$We have, \cos \frac{π}{65} \cos \frac{2 π}{65} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65} = \frac{2 \sin \frac{π}{65}}{2 \sin \frac{π}{65}} \cos \frac{π}{65} \cos \frac{2 π}{65} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65} (dividing and multiplying by 2 \sin \frac{π}{65}) = \frac{2 \sin \frac{2 π}{65}}{2 \times 2 \sin \frac{π}{65}} \cos \frac{2 π}{65} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65} = \frac{2 \sin \frac{4 π}{65}}{2 \times 4 \sin \frac{π}{65}} \cos \frac{4 π}{65} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65}$
$= \frac{2 \sin \frac{8 π}{65}}{2 \times 8 \sin \frac{π}{65}} \cos \frac{8 π}{65} \cos \frac{16 π}{65} \cos \frac{32 π}{65} = \frac{2 \sin \frac{16 π}{65}}{2 \times 16 \sin \frac{π}{65}} \cos \frac{16 π}{65} \cos \frac{32 π}{65} = \frac{2 \sin \frac{32 π}{65}}{2 \times 32 \sin \frac{π}{65}} \cos \frac{32 π}{65} = \frac{\sin \frac{64 π}{65}}{64 \sin \frac{π}{65}} = \frac{\sin (π - \frac{π}{65})}{64 \sin \frac{π}{65}} = \frac{\sin \frac{π}{65}}{64 \sin \frac{π}{65}} = \frac{1}{64}$

Page No 9.43:

Question 4:

If $\cos 2 x + 2 \cos x = 1$ then, $(2 - \cos^{2} x) \sin^{2} x$ is equal to
(a) 1

(b) $- 1$

(c) $- \sqrt{5}$

(d) $\sqrt{5}$

Answer:

(a) 1

$We have, \cos 2 x + 2 \cos x = 1 \Rightarrow 2 \cos^{2} x - 1 + 2 \cos x = 1 \Rightarrow \cos^{2} x + \cos x - 1 = 0 \Rightarrow \cos x = \frac{- 1 \pm \sqrt{1^{2} + 4}}{2} \Rightarrow \cos x = \frac{- 1 \pm \sqrt{5}}{2} \Rightarrow \cos x = \frac{- 1 + \sqrt{5}}{2}$

$Now, (2 - \cos^{2} x) \sin^{2} x = [2 - {(\frac{- 1 + \sqrt{5}}{2})}^{2}] (1 - \cos^{2} x) = [2 - \frac{1}{4} (1 - 2 \sqrt{5} + 5)] (1 - \frac{1}{4} (1 - 2 \sqrt{5} + 5)) = \frac{1}{4} (1 + \sqrt{5}) (\sqrt{5} - 1) = \frac{4}{4} = 1$

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Question 5:

For all real values of x, $\cot x - 2 \cot 2 x$ is equal to
(a) $\tan 2 x$
(b) $\tan x$
(c) $- \cot 3 x$
(d) none of these

Answer:

(b) $\tan x$

$We have, \cot x - 2 \cot 2 x = \cot x - 2 \frac{\cot^{2} x - 1}{2 \cot x} = \frac{\cot^{2} x - \cot^{2} x + 1}{\cot x} = \frac{1}{\cot x} = \tan x$

Page No 9.43:

Question 6:

The value of $2 \tan \frac{π}{10} + 3 \sec \frac{π}{10} - 4 \cos \frac{π}{10}$ is
(a) 0
(b) $\sqrt{5}$
(c) 1
(d) none of these

Answer:

(a) 0

$We have, 2 \tan \frac{π}{10} + 3 \sec \frac{π}{10} - 4 \cos \frac{π}{10} = 2 \tan 18 ° + 3 \sec 18 ° - 4 \cos 18 ° = 2 \frac{\sin 18 °}{\cos 18 °} + 3 \times \frac{1}{\cos 18 °} - 4 \cos 18 ° = 2 \times \frac{\frac{\sqrt{5} - 1}{4}}{\frac{\sqrt{10 + 2 \sqrt{5}}}{4}} + 3 \times \frac{1}{\frac{\sqrt{10 + 2 \sqrt{5}}}{4}} - 4 \times \frac{\sqrt{10 + 2 \sqrt{5}}}{4} = 2 \times \frac{\sqrt{5} - 1}{\sqrt{10 + 2 \sqrt{5}}} + 3 \times \frac{4}{\sqrt{10 + 2 \sqrt{5}}} - \sqrt{10 + 2 \sqrt{5}} = \frac{2 \sqrt{5} - 2 + 12 - {(\sqrt{10 + 2 \sqrt{5}})}^{2}}{(\sqrt{10 + 2 \sqrt{5}})} = \frac{2 \sqrt{5} + 10 - 10 - 2 \sqrt{5}}{(\sqrt{10 + 2 \sqrt{5}})} = 0$

Page No 9.43:

Question 7:

If in a $∆ A B C, \tan A + \tan B + \tan C = 0$ , then $\cot A \cot B \cot C =$
(a) 6
(b) 1
(c) $\frac{1}{6}$
(d) none of these

Answer:

(d) none of these

ABC is a triangle.

$∴ A + B + C = π \Rightarrow A + B = π - C \Rightarrow \tan (A + B) = \tan (π - C) \Rightarrow \frac{\tan A + \tan B}{1 - \tan A \tan B} = - \tan C \Rightarrow \tan A + \tan B = - \tan C + \tan A \tan B \tan C \Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \tan C \Rightarrow 0 = \tan A \tan B \tan C [Given : \tan A \tan B \tan C = 0] \Rightarrow \tan A \tan B \tan C = 0 \Rightarrow \frac{1}{\tan A \tan B \tan C} = \frac{1}{0} \Rightarrow \cot A \cot B \cot C \to \infty$

Page No 9.43:

Question 8:

If $\cos x = \frac{1}{2} (a + \frac{1}{a}),$ and $\cos 3 x = λ (a^{3} + \frac{1}{a^{3}})$ , then $λ =$
(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) 1

(d) none of these

Answer:

(b) $\frac{1}{2}$
$Given : \cos x = \frac{1}{2} (a + \frac{1}{a}) \cos 3 x = λ (a^{3} + \frac{1}{a^{3}}) Now, \cos^{3} x = \frac{1}{8} [a^{3} + \frac{1}{a^{3}} + 3 a \frac{1}{a} (a + \frac{1}{a})] \Rightarrow \cos^{3} x = \frac{1}{8} (a^{3} + \frac{1}{a^{3}} + 3 \times 2 \cos x) [∵ \cos x = \frac{1}{2} (a + \frac{1}{a})] \Rightarrow \cos^{3} x = \frac{1}{8} (\frac{\cos 3 x}{λ} + 6 \cos x) \Rightarrow \cos^{3} x = \frac{1}{8} (\frac{4 \cos^{3} x - 3 \cos x}{λ} + 6 \cos x) \Rightarrow \cos^{3} x = \frac{4 \cos^{3} x}{8 λ} - \frac{3 \cos x}{8 λ} + \frac{6 \cos x}{8} On comparing the powers of \cos^{3} x on both sides, we get 1 = \frac{4}{8 λ} \Rightarrow λ = \frac{1}{2}$

Page No 9.43:

Question 9:

If $2 \tan α = 3 \tan β, then \tan (α - β) =$

(a) $\frac{\sin 2 β}{5 - \cos 2 β}$

(b) $\frac{\cos 2 β}{5 - \cos 2 β}$

(c) $\frac{\sin 2 β}{5 + \cos 2 β}$

(d) none of these

Answer:

(a) $\frac{\sin 2 β}{5 - \cos 2 β}$

$Given : 2 \tan α = 3 \tan β Now, \tan (α - β) = \frac{tanα - tanβ}{1 + tanαtanβ} = \frac{\frac{3}{2} tanβ - tanβ}{1 + (\frac{3}{2} tanβ) tanβ} = \frac{3 tanβ - 2 tanβ}{2 + 3 \tan^{2} β} = \frac{tanβ}{2 + 3 \tan^{2} β} = \frac{\frac{sinβ}{cosβ}}{2 + 3 \frac{\sin^{2} β}{\cos^{2} β}} = \frac{sinβcosβ}{2 \cos^{2} β + 3 \sin^{2} β} = \frac{sinβcosβ}{2 + \sin^{2} β} = \frac{2 sinβcosβ}{4 + 2 \sin^{2} β} = \frac{\sin 2 β}{4 + 1 - \cos 2 β} ∴ \tan (α - β) = \frac{\sin 2 β}{5 - \cos 2 β}$

Page No 9.43:

Question 10:

If $\tan α = \frac{1 - \cos β}{\sin β}$ , then

(a) $\tan 3 α = \tan 2 β$

(b) $\tan 2 α = \tan β$

(c) $\tan 2 β = \tan α$

(d) none of these

Answer:

(b) $\tan 2 α = \tan β$

$tanα = \frac{1 - cosβ}{sinβ} = \frac{2 \sin^{2} \frac{β}{2}}{2 \sin \frac{β}{2} \cos \frac{β}{2}} = \frac{\sin \frac{β}{2}}{\cos \frac{β}{2}} \Rightarrow tanα = \tan \frac{β}{2} \Rightarrow α = \frac{β}{2} \Rightarrow 2 α = β ∴ \tan 2 α = tanβ$

Page No 9.43:

Question 11:

If $\sin α + \sin β = a and \cos α - \cos β = b then \tan \frac{α - β}{2} =$
(a) $- \frac{a}{b}$

(b) $- \frac{b}{a}$

(c) $\sqrt{a^{2} + b^{2}}$

(d) none of these

Answer:

(b) $- \frac{b}{a}$

$Given : sinα + sinβ = a \Rightarrow 2 \sin \frac{α + β}{2} \cos \frac{α - β}{2} = a . . . (1) Also, cosα + cosβ = b \Rightarrow - 2 \sin \frac{α + β}{2} \sin \frac{α - β}{2} = b . . . (2) On dividing (1) by (2), we get \frac{- \cos \frac{α - β}{2}}{\sin \frac{α - β}{2}} = \frac{a}{b} \Rightarrow \frac{- \sin \frac{α - β}{2}}{\cos \frac{α - β}{2}} = \frac{b}{a} \Rightarrow \tan \frac{α - β}{2} = - \frac{b}{a}$

Page No 9.43:

Question 12:

The value of ${(\cot \frac{x}{2} - \tan \frac{x}{2})}^{2} (1 - 2 \tan x \cot 2 x)$ is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(d) 4

$We have, {(\cot \frac{x}{2} - \tan \frac{x}{2})}^{2} (1 - 2 \tan x \cot 2 x) (\cot^{2} \frac{x}{2} - 2 \cot \frac{x}{2} \tan \frac{x}{2} + \tan^{2} \frac{x}{2}) \{1 - 2 \tan x (\frac{\cot^{2} x - 1}{2 \cot x})\} (\cot^{2} \frac{x}{2} - 2 + \tan^{2} \frac{x}{2}) \{1 - \tan x (\frac{\cot^{2} x - 1}{\cot x})\} (\cot^{2} \frac{x}{2} + \tan^{2} \frac{x}{2} - 2) (1 - \frac{\cot x - \tan x}{\cot x}) (\cot^{2} \frac{x}{2} + \tan^{2} \frac{x}{2} - 2) (\tan^{2} x) (\cot^{2} \frac{x}{2} + \tan^{2} \frac{x}{2} - 2) {(\frac{2 \tan \frac{x}{2}}{1 - \tan^{2} \frac{x}{2}})}^{2}$
$= \frac{1}{{(1 - \tan^{2} \frac{x}{2})}^{2}} (4 + 4 \tan^{4} \frac{x}{2} - 8 \tan^{2} \frac{x}{2}) = \frac{1}{{(1 - \tan^{2} \frac{x}{2})}^{2}} (4 - 8 \tan^{2} \frac{x}{2} + 4 \tan^{4} \frac{x}{2}) = \frac{4}{{(1 - \tan^{2} \frac{x}{2})}^{2}} \{{(\tan^{2} \frac{x}{2})}^{2} - 2 (\tan^{2} \frac{x}{2}) + 1\} = \frac{4 {(\tan^{2} \frac{x}{2} - 1)}^{2}}{{(1 - \tan^{2} \frac{x}{2})}^{2}} = 4$

Page No 9.43:

Question 13:

The value of $\tan x \sin (\frac{π}{2} + x) \cos (\frac{π}{2} - x)$
(a) 1
(b) $- 1$
(c) $\frac{1}{2} \sin 2 x$
(d) none of these.

Answer:

(d) none of these

$We have, tanθ \sin (\frac{π}{2} + x) \cos (\frac{π}{2} - x) = \frac{\sin x}{\cos x} \cos x \sin x = \sin^{2} x$

Page No 9.44:

Question 14:

$\sin^{2} (\frac{π}{18}) + \sin^{2} (\frac{π}{9}) + \sin^{2} (\frac{7 π}{18}) + \sin^{2} (\frac{4 π}{9}) =$

(a) 1
(b) 2
(c) 4
(d) none of these.

Answer:

(b) 2

$We have, \sin^{2} (\frac{π}{18}) + \sin^{2} (\frac{π}{9}) + \sin^{2} (\frac{7 π}{18}) + \sin^{2} (\frac{4 π}{9}) = \frac{1}{2} [1 - \cos (\frac{π}{9}) + 1 - \cos (\frac{2 π}{9}) + 1 - \cos \frac{7 π}{9} + 1 - \cos \frac{8 π}{9}] (∵ \sin^{2} θ = \frac{1 - \cos 2 θ}{2}) = \frac{1}{2} [4 - \cos (\frac{π}{9}) - \cos (\frac{2 π}{9}) - \{- \cos (π - \frac{7 π}{9})\} - \{- \cos (π - \frac{8 π}{9})\}] = \frac{1}{2} [4 - \cos (\frac{π}{9}) - \cos (\frac{2 π}{9}) + \cos (\frac{2 π}{9}) + \cos (\frac{π}{9})] = \frac{4}{2} = 2$

Page No 9.44:

Question 15:

If $5 \sin α = 3 \sin (α + 2 β) \neq 0$ , then $\tan (α + β)$ is equal to
(a) $2 \tan β$
(b) $3 \tan β$
(c) $4 \tan β$
(d) $6 \tan β$

Answer:

(c) $4 \tan β$

$We have, 5 \sin α = 3 \sin (α + 2 β) \Rightarrow \frac{5}{3} = \frac{\sin (α + 2 β)}{\sin α} \Rightarrow \frac{5 - 3}{5 + 3} = \frac{\sin (α + 2 β) - \sin α}{\sin (α + 2 β) + \sin α} (Using componendo and d ividendo) \Rightarrow \frac{2}{8} = \frac{\sin (α + 2 β) - \sin α}{\sin (α + 2 β) + \sin α} \Rightarrow \frac{1}{4} = \frac{2 \cos \frac{α + 2 β + α}{2} \sin \frac{α + 2 β - α}{2}}{2 \sin \frac{α + 2 β + α}{2} \cos \frac{α + 2 β - α}{2}} \Rightarrow \frac{1}{4} = \frac{\cos (α + β) \sin β}{\sin (α + β) \cos β} \Rightarrow \frac{1}{4} = \cot (α + β) \tan β \Rightarrow \frac{1}{4} = \frac{1}{\tan (α + β)} \tan β ∴ \tan (α + β) = 4 \tan β$

Page No 9.44:

Question 16:

The value of $2 \cos x - \cos 3 x - \cos 5 x - 16 \cos^{3} x \sin^{2} x$ is
(a) 2
(b) 1
(c) 0
(d) −1

Answer:

(c) 0

$We have, 2 \cos x - \cos 3 x - \cos 5 x - 16 \cos^{3} x \sin^{2} x = 2 \cos x - \cos 3 x - \cos 5 x - 16 [\frac{\cos 3 x + 3 \cos x}{4} \times \frac{(1 - \cos 2 x)}{2}] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [(\cos 3 x + 3 \cos x) (1 - \cos 2 x)] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [\cos 3 x - \cos 3 x \cos 2 x + 3 \cos x - 3 \cos x \cos 2 x] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [\cos 3 x + 3 \cos x] + 2 \cos 3 x \cos 2 x + 3 [2 \cos x \cos 2 x] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [\cos 3 x + 3 \cos x] + \cos 5 x + \cos x + 3 \cos 3 x + 3 \cos x [2 cosAcosB = \cos (A + B) + \cos (A - B)] = 2 \cos x - \cos 3 x - \cos 5 x - 2 \cos 3 x - 6 \cos x + \cos 5 x + \cos x + 3 \cos 3 x + 3 \cos x = 0$

Page No 9.44:

Question 17:

If $A = 2 \sin^{2} x - \cos 2 x$ , then A lies in the interval
(a) $[- 1, 3]$
(b) $[1, 2]$
(c) $[- 2, 4]$
(d) none of these

Answer:

(a) $[- 1, 3]$

$A = 2 \sin^{2} x - \cos 2 x = 2 \sin^{2} x - (1 - 2 \sin^{2} x) = 4 \sin^{2} x - 1 ∵ 0 \leq \sin^{2} x \leq 1 \Rightarrow 4 \times 0 \leq 4 \times \sin^{2} x \leq 4 \times 1 \Rightarrow 0 \leq 4 \sin^{2} x \leq 4 \Rightarrow 0 - 1 \leq 4 \sin^{2} x - 1 \leq 4 - 1 \Rightarrow - 1 \leq 4 \sin^{2} x - 1 \leq 3 \Rightarrow - 1 \leq A \leq 3 \Rightarrow A \in [- 1, 3]$

Page No 9.44:

Question 18:

The value of $\frac{\cos 3 x}{2 \cos 2 x - 1}$ is equal to
(a) cos x
(b) sin x
(c) tan x
(d) none of these

Answer:

(a) cos x

$We have, ∴ \frac{\cos 3 x}{2 \cos 2 x - 1} = \frac{4 \cos^{3} x - 3 \cos x}{2 (2 \cos^{2} x - 1) - 1} [∵ \cos 3 x = 4 \cos^{3} x - 3 \cos x] = \frac{4 \cos^{3} x - 3 \cos x}{4 \cos^{2} x - 2 - 1} = \frac{4 \cos^{3} x - 3 \cos x}{4 \cos^{2} x - 3} = \cos x (\frac{4 \cos^{2} x - 3}{4 \cos^{2} x - 3}) = \cos x$

Page No 9.44:

Question 19:

If $\tan (π / 4 + x) + \tan (π / 4 - x) = λ \sec 2 x, then$
(a) 3
(b) 4
(c) 1
(d) 2

Answer:

(d) 2

$Given : \tan (\frac{π}{4} + x) + \tan (\frac{π}{4} - x) = λ \sec 2 x \Rightarrow \frac{\tan \frac{π}{4} + \tan x}{1 - \tan \frac{π}{4} \times \tan x} + \frac{\tan \frac{π}{4} - \tan x}{1 + \tan \frac{π}{4} \times \tan x} = λ \sec 2 x \Rightarrow \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} = λ \sec 2 x \Rightarrow \frac{{(1 + \tan x)}^{2} + {(1 - \tan x)}^{2}}{(1 - \tan x) (1 + \tan x)} = λ \sec 2 x \Rightarrow \frac{2 (1 + \tan^{2} x)}{1 - \tan^{2} x} = λ \sec 2 x$

$\Rightarrow \frac{2 \sec^{2} x}{1 - \tan^{2} x} = λ \sec 2 x \Rightarrow \frac{2}{\cos^{2} x (1 - \tan^{2} x)} = λ \sec 2 x \Rightarrow \frac{2}{\cos^{2} x (1 - \frac{\sin^{2} x}{\cos^{2} x})} = λ \sec 2 x \Rightarrow \frac{2}{\cos^{2} x - \sin^{2} x} = λ \sec 2 x \Rightarrow \frac{2}{\cos 2 x} = λ \sec 2 x \Rightarrow 2 \sec 2 x = λ \sec 2 x \Rightarrow 2 = λ ∴ λ = 2$

Page No 9.44:

Question 20:

The value of $\cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x)$ is
(a) $\frac{1}{2} \cos 2 x$

(b) 0

(c) $- \frac{1}{2} \cos 2 x$

(d) $\frac{1}{2}$

Answer:

(a) $\frac{1}{2} \cos 2 x$

$We have, \cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x) = \cos^{2} (\frac{π}{6} + x) - \cos^{2} [\frac{π}{2} - (\frac{π}{6} - x)] = \cos^{2} (\frac{π}{6} + x) - \cos^{2} (\frac{π}{3} + x) = [\cos (\frac{π}{6} + x) + \cos (\frac{π}{3} + x)] [\cos (\frac{π}{6} + x) - \cos (\frac{π}{3} + x)] = 2 \cos (\frac{\frac{π}{6} + x + \frac{π}{3} + x}{2}) \cos (\frac{\frac{π}{6} + x - \frac{π}{3} - x}{2}) 2 \sin (\frac{\frac{π}{6} + x + \frac{π}{3} + x}{2}) \sin (\frac{\frac{π}{3} + x - \frac{π}{6} - x}{2}) = 4 \cos (\frac{π}{4} + x) \cos (- \frac{π}{12}) \sin (\frac{π}{4} + x) \sin (\frac{π}{12}) = 4 \cos (\frac{π}{4} + x) \cos (\frac{π}{12}) \sin (\frac{π}{4} + x) \sin (\frac{π}{12}) = [2 \sin (\frac{π}{4} + x) \cos (\frac{π}{4} + x)] [2 \sin (\frac{π}{12}) \cos (\frac{π}{12})] = \sin (\frac{π}{2} + 2 x) \sin \frac{π}{6} = \cos 2 x \times \frac{1}{2} = \frac{1}{2} \cos 2 x$

Page No 9.44:

Question 21:

$\frac{\sin 3 x}{1 + 2 \cos 2 x}$ is equal to

(a) cos x
(b) sin x
(c) – cos x
(d) sin x

Answer:

(b) sin x

$We have, \frac{\sin 3 x}{1 + 2 \cos 2 x} = \frac{3 \sin x - 4 \sin^{3} x}{1 + 2 (1 - 2 \sin^{2} x)} = \frac{3 \sin x - 4 \sin^{3} x}{1 + 2 - 4 \sin^{2} x} = \frac{\sin x (3 - 4 \sin^{2} x)}{(3 - 4 \sin^{2} x)} = \sin x$

Page No 9.44:

Question 22:

The value of $2 \sin^{2} B + 4 \cos (A + B) \sin A \sin B + \cos 2 (A + B)$ is
(a) 0
(b) cos 3A
(c) cos 2A
(d) none of these

Answer:

(c) cos 2A

$We have, 2 \sin^{2} B + 4 \cos (A + B) \sin A \sin B + \cos 2 (A + B) = 1 - \cos 2 B + \cos 2 (A + B) + 4 \cos (A + B) \sin A \sin B = 1 + (\cos 2 (A + B) - \cos 2 B) + 4 \cos (A + B) \sin A \sin B = 1 - 2 \sin A \sin (A + 2 B) + 4 \cos (A + B) \sin A \sin B [∵ \cos C - \cos D = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}] = 1 - 2 \sin A [\sin (A + 2 B) - 2 \sin B \cos (A + B)] = 1 - 2 \sin A [\sin (A + 2 B) - \{\sin (B + A + B) + \sin (B - (A + B))\}] [∵ 2 \sin C \cos D = \sin (C + D) + \sin (C - D)] = 1 - 2 \sin A [\sin (A + 2 B) - \{\sin (A + 2 B) + \sin (- A)\}] = 1 - 2 \sin A [\sin A] = 1 - 2 \sin^{2} A = \cos 2 A$

Page No 9.44:

Question 23:

The value of $\frac{2 (\sin 2 x + 2 \cos^{2} x - 1)}{\cos x - \sin x - \cos 3 x + \sin 3 x}$ is
(a) cos x
(b) sec x
(c) cosec x
(d) sin x

Answer:

(c) cosec x

$We have, \frac{2 (\sin 2 x + 2 \cos^{2} x - 1)}{\cos x - \sin x - \cos 3 x + \sin 3 x} = \frac{2 (\sin 2 x + \cos 2 x)}{\cos x - \sin x - 4 \cos^{3} x + 3 \cos x + 3 \sin x - 4 \sin^{3} x} = \frac{2 (\sin 2 x + \cos 2 x)}{4 \cos x - 4 \cos^{3} x + 2 \sin x - 4 \sin^{3} x} = \frac{2 (\sin 2 x + \cos 2 x)}{4 \cos x (1 - \cos^{2} x) + 2 \sin x (1 - 2 \sin^{2} x)} = \frac{2 (\sin 2 x + \cos 2 x)}{4 \cos x \sin^{2} x + 2 \sin x \cos 2 x} = \frac{2 (\sin 2 x + \cos 2 x)}{2 \times 2 \sin x \cos x \sin x + 2 \sin x \cos 2 x} = \frac{2 (\sin 2 x + \cos 2 x)}{2 \sin 2 x \sin x + 2 \sin x \cos 2 x} = \frac{2 (\sin 2 x + \cos 2 x)}{2 \sin x (\sin 2 x + \cos 2 x)} = \frac{1}{\sin x} = cosec x$

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Question 24:

$2 (1 - 2 \sin^{2} 7 x) \sin 3 x$ is equal to
(a) $\sin 17 x - \sin 11 x$
(b) $\sin 11 x - \sin 17 x$
(c) $\cos 17 x - \cos 11 x$
(d) $\cos 17 x + \cos 11 x$

Answer:

(a) $\sin 17 x - \sin 11 x$

$We have, 2 (1 - 2 \sin^{2} 7 x) \sin 3 x = 2 (\cos 14 x) \sin 3 x [∵ \cos 2 x = 1 - 2 \sin^{2} x] = 2 \sin 3 x \cos 14 x = \sin 17 x - \sin 11 x [∵ 2 sinA cosB = \sin (A + B) - \sin (A - B)] ∴ 2 (1 - 2 \sin^{2} 7 x) \sin 3 x = \sin 17 x - \sin 11 x$

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Question 25:

If α and β are acute angles satisfying $\cos 2 α = \frac{3 \cos 2 β - 1}{3 - \cos 2 β}$ , then tan α =
(a) $\sqrt{2} \tan β$

(b) $\frac{1}{\sqrt{2}} \tan β$

(c) $\sqrt{2} \cot β$

(d) $\frac{1}{\sqrt{2}} \cot β$

Answer:

(a) $\sqrt{2} \tan β$

$Given : \cos 2 α = \frac{3 \cos 2 β - 1}{3 - \cos 2 β} \Rightarrow \frac{\cos 2 α - 1}{\cos 2 α + 1} = \frac{(3 \cos 2 β - 1) - (3 - \cos 2 β)}{(3 \cos 2 β - 1) + (3 - \cos 2 β)} (Using componendo and d ividendo) \Rightarrow \frac{\cos 2 α - 1}{\cos 2 α + 1} = \frac{4 \cos 2 β - 4}{2 \cos 2 β + 2} \Rightarrow - \frac{1 - \cos 2 α}{1 + \cos 2 α} = \frac{- 4 (1 - \cos 2 β)}{2 (1 + \cos 2 β)} \Rightarrow \frac{1 - \cos 2 α}{1 + \cos 2 α} = \frac{2 (1 - \cos 2 β)}{(1 + \cos 2 β)} \Rightarrow \frac{2 \sin^{2} α}{2 \cos^{2} α} = \frac{2 (2 \sin^{2} β)}{(2 \cos^{2} β)} \Rightarrow \tan^{2} α = 2 \tan^{2} β ∴ \tan α = \sqrt{2} \tan β$

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Question 26:

If $\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{α}{2}$ , then $\cos α =$

(a) $1 - e \cos (\cos x + e)$

(b) $\frac{1 + e \cos x}{\cos x - e}$

(c) $\frac{1 - e \cos x}{\cos x - e}$

(d) $\frac{\cos x - e}{1 - e \cos x}$

Answer:

(d) $\frac{\cos x - e}{1 - e \cos x}$

$Given : \tan \frac{x}{2} = \sqrt{\frac{1 - e}{1 + e}} \tan \frac{α}{2} \Rightarrow \frac{\tan \frac{x}{2}}{\tan \frac{α}{2}} = \sqrt{\frac{1 - e}{1 + e}} Squaring both sides, we get, \frac{\tan^{2} \frac{x}{2}}{\tan^{2} \frac{α}{2}} = \frac{1 - e}{1 + e} \Rightarrow \tan^{2} \frac{α}{2} (1 - e) = \tan^{2} \frac{x}{2} (1 + e)$

$\Rightarrow \frac{\sin^{2} \frac{α}{2}}{\cos^{2} \frac{α}{2}} (1 - e) = \frac{\sin^{2} \frac{x}{2}}{\cos^{2} \frac{x}{2}} (1 + e) \Rightarrow \frac{\frac{1}{2} (1 - cosα)}{\frac{1}{2} (1 + cosα)} (1 - e) = \frac{\frac{1}{2} (1 - \cos x)}{\frac{1}{2} (1 + \cos x)} (1 + e) \Rightarrow (1 - cosα) (1 + \cos x) (1 - e) = (1 + cosα) (1 - \cos x) (1 + e) \Rightarrow (1 + \cos x) (1 - e) - cosα (1 + \cos x) (1 - e) = (1 - \cos x) (1 + e) + cosα (1 - \cos x) (1 + e) \Rightarrow cosα \{(1 + \cos x) (1 - e) + (1 - \cos x) (1 + e)\} = (1 + \cos x) (1 - e) - (1 - \cos x) (1 + e) \Rightarrow cosα = \frac{2 \cos x - 2 e}{2 - 2 ecos x} = \frac{\cos x - e}{1 - ecos x}$

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Question 27:

If $(2^{n} + 1) x = π,$ then $2^{n} \cos x \cos 2 x \cos 2^{2} x . . . \cos 2^{n - 1} x = 1$
(a) $- 1$
(b) 1
(c) 1/2
(d) None of these

Answer:

(b) 1

$(2^{n} + 1) x = π (Given) \Rightarrow 2^{n} x + x = π \Rightarrow 2^{n} x = π - x \Rightarrow \sin 2^{n} x = \sin (π - x) \Rightarrow \sin 2^{n} x = \sin x . . . (1)$

$2^{n} \cos x \cos 2 x \cos 2^{2} x . . . \cos 2^{n - 1} x = 2^{n} \times \frac{\sin 2^{n} x}{2^{n} \sin x} = \frac{\sin 2^{n} x}{\sin x} = \frac{\sin x}{\sin x} [From (1)] = 1$

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Question 28:

If $\tan x = t$ then $\tan 2 x + \sec 2 x =$

(a) $\frac{1 + t}{1 - t}$

(b) $\frac{1 - t}{1 + t}$

(c) $\frac{2 t}{1 - t}$

(d) $\frac{2 t}{1 + t}$

Answer:

(a) $\frac{1 + t}{1 - t}$

$\tan 2 x + \sec 2 x = \frac{2 \tan x}{1 - \tan^{2} x} + \frac{1 + \tan^{2} x}{1 - \tan^{2} x} = \frac{2 \tan x + 1 + \tan^{2} x}{1 - \tan^{2} x} = \frac{{(1 + \tan x)}^{2}}{1 - \tan^{2} x} = \frac{(1 + \tan x) (1 + \tan x)}{(1 + \tan x) (1 - \tan x)} = \frac{1 + \tan x}{1 - \tan x} = \frac{1 + t}{1 - t} [\tan x = t (given)]$

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Question 29:

The value of $\cos^{4} x + \sin^{4} x - 6 \cos^{2} x \sin^{2} x$ is
(a) cos 2x
(b) sin 2x
(c) cos 4x
(d) none of these

Answer:

(c) cos 4x

$\cos^{4} x + \sin^{4} x - 6 \cos^{2} x \sin^{2} x = \cos^{4} x + \sin^{4} x - 2 \cos^{2} x \sin^{2} x - 4 \cos^{2} x \sin^{2} x = {(\cos^{2} x - \sin^{2} x)}^{2} - {(2 \sin x \cos x)}^{2} = \cos^{2} 2 x - \sin^{2} 2 x = \cos 4 x$

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Question 30:

The value of $\cos (36 ° - A) \cos (36 ° + A) + \cos (54 ° - A) \cos (54 ° + A)$ is
(a) cos 2A
(b) sin 2A
(c) cos A
(d) 0

Answer:

(a) cos 2A

$\cos (36 ° - A) \cos (36 ° + A) + \cos (54 ° - A) \cos (54 ° + A) = \cos [90 ° - (54 ° + A)] \cos [90 ° - (54 ° - A)] + \cos (54 ° - A) \cos (54 ° + A) = \sin (54 ° + A) \sin (54 ° - A) + \cos (54 ° - A) \cos (54 ° + A) [\cos (90 ° - θ) = \sin θ] = \cos (54 ° + A - 54 ° + A) [\cos (A - B) = cosAcosB + sinAsinB] = \cos 2 A$

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Question 31:

The value of $\tan x \tan (\frac{π}{3} - x) \tan (\frac{π}{3} + x)$ is
(a) cot 3x
(b) 2cot 3x
(c) tan 3x
(d) 3 tan 3x

Answer:

(c) tan 3x

$\frac{π}{3} = 60 ° \tan x \tan (60 ° - x) \tan (60 ° + x) = \tan x \times \frac{\tan 60 ° - \tan x}{1 + \tan 60 ° \tan x} \times \frac{\tan 60 ° + \tan x}{1 - \tan 60 ° \tan x} = \tan x \times \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} \times \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} = \frac{\tan x (3 - \tan^{2} x)}{1 - 3 \tan^{2} x} = \frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x} = \tan 3 x$

Page No 9.45:

Question 32:

The value of $\tan x + \tan (\frac{π}{3} + x) + \tan (\frac{2 π}{3} + x)$ is
(a) 3 tan 3x
(b) tan 3x
(c) 3 cot 3x
(d) cot 3x

Answer:

(a) 3 tan 3x

$\frac{π}{3} = 60 °, \frac{2 π}{3} = 120 ° \tan x + \tan (60 ° + x) + \tan (120 ° + x) = \tan x + \frac{\tan 60 ° + \tan x}{1 - \tan 60 ° \tan x} + \frac{\tan 120 ° + \tan x}{1 - \tan 120 ° \tan x} = \tan x + \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} + \frac{(- \sqrt{3} + \tan x)}{1 + \sqrt{3} \tan x} = \frac{\tan x (1 - 3 \tan^{2} x) + (\sqrt{3} + \tan x) (1 + \sqrt{3} \tan x) + (- \sqrt{3} + \tan x) (1 - \sqrt{3} \tan x)}{1 - 3 \tan^{2} x} = \frac{\tan x - 3 \tan^{3} x + \sqrt{3} + 3 \tan x + \tan x + \sqrt{3} \tan^{2} x + \tan x - \sqrt{3} \tan^{2} x - \sqrt{3} + 3 \tan x}{1 - 3 \tan^{2} x} = \frac{9 \tan x - 3 \tan^{3} x}{1 - 3 \tan^{2} x} = \frac{3 (3 \tan x - \tan^{3} x)}{1 - 3 \tan^{2} x} = 3 \tan 3 x$

Page No 9.45:

Question 33:

The value of $\frac{\sin 5 α - \sin 3 α}{\cos 5 α + 2 \cos 4 α + \cos 3 α} =$
(a) $\cot α / 2$
(b) $\cot α$
(c) $\tan α / 2$
(d) None of these

Answer:

(c) $\tan α / 2$

$\frac{\sin 5 α - \sin 3 α}{\cos 5 α + 2 \cos 4 α + \cos 3 α} = \frac{\sin 5 α - \sin 3 α}{\cos 5 α + \cos 3 α + 2 \cos 4 α} = \frac{2 \sin α \cos 4 α}{2 \cos 4 α \cos α + 2 \cos 4 α} = \frac{2 \sin α \cos 4 α}{2 \cos 4 α (\cos α + 1)} = \frac{\sin α}{\cos α + 1} = \frac{2 \sin \frac{α}{2} \cos \frac{α}{2}}{\cos^{2} \frac{α}{2} - \sin^{2} \frac{α}{2} + \sin^{2} \frac{α}{2} + \cos^{2} \frac{α}{2}} = \frac{2 \sin \frac{α}{2} \cos \frac{α}{2}}{2 \cos^{2} \frac{α}{2}} = \frac{\sin \frac{α}{2}}{\cos \frac{α}{2}} = \tan \frac{α}{2}$

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Question 34:

$\frac{\sin 5 x}{\sin x}$ is equal to
(a) $16 \cos^{4} x - 12 \cos^{2} x + 1$
(b) $16 \cos^{4} x + 12 \cos^{2} x + 1$
(c) $16 \cos^{4} x - 12 \cos^{2} x - 1$
(d) $16 \cos^{4} x + 12 \cos^{2} x - 1$

Answer:

(a) $16 \cos^{4} x - 12 \cos^{2} x + 1$

$To find : \frac{\sin 5 x}{\sin x} Now, \sin 5 x = \sin (3 x + 2 x) = \sin 3 x \cos 2 x + \cos 3 x \sin 2 x = (3 \sin x - 4 \sin^{3} x) (1 - 2 \sin^{2} x) + (4 \cos^{3} x - 3 \cos x) (2 \sin x \cos x) = (3 \sin x - 6 \sin^{3} x - 4 \sin^{3} x + 8 \sin^{5} x) + 2 \sin x \cos^{2} x (4 \cos^{2} x - 3) = (3 \sin x - 10 \sin^{3} x + 8 \sin^{5} x) + 2 \sin x (1 - \sin^{2} x) [4 (1 - \sin^{2} x) - 3] = (3 \sin x - 10 \sin^{3} x + 8 \sin^{5} x) + (2 \sin x - 2 \sin^{3} x) (4 - 4 \sin^{2} x - 3) = (3 \sin x - 10 \sin^{3} x + 8 \sin^{5} x) + (2 \sin x - 8 \sin^{3} x - 2 \sin^{3} x + 8 \sin^{5} x) = 5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x ∴ \frac{\sin 5 x}{\sin x} = \frac{5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x}{\sin x} = 5 - 20 \sin^{2} x + 16 \sin^{4} x = 5 - 20 (1 - \cos^{2} x) + 16 {(1 - \cos^{2} x)}^{2} = 5 - 20 + 20 \cos^{2} x + 16 (1 + \cos^{4} x - 2 \cos^{2} x) = 5 - 20 + 20 \cos^{2} x + 16 + 16 \cos^{4} x - 32 \cos^{2} x = 16 \cos^{4} x - 12 \cos^{2} x + 1$

Page No 9.45:

Question 35:

If $n = 1, 2, 3, . . ., then \cos α \cos 2 α \cos 4 α . . . \cos 2^{n - 1} α$ is equal to

(a) $\frac{\sin 2 n α}{2 n \sin α}$

(b) $\frac{\sin 2^{n} α}{2^{n} \sin 2^{n - 1} α}$

(c) $\frac{\sin 4^{n - 1} α}{4^{n - 1} \sin α}$

(d) $\frac{\sin 2^{n} α}{2^{n} \sin α}$

(e) None of these

Answer:

(d) $\frac{\sin 2^{n} α}{2^{n} \sin α}$

$∵ \cos α \cos 2 α \cos 4 α . . . \cos 2^{n - 1} α = \frac{\sin 2^{n} α}{2^{n} \sin α}$

Page No 9.45:

Question 36:

If $\tan x = \frac{a}{b}$ , then $b \cos 2 x + a \sin 2 x$ is equal to

(a) a

(b) b

(c) $\frac{a}{b}$

(d) $\frac{b}{a}$

Answer:

Given: $\tan x = \frac{a}{b}$
Now,

$b \cos 2 x + a \sin 2 x = b (\frac{1 - \tan^{2} x}{1 + \tan^{2} x}) + a (\frac{2 \tan x}{1 + \tan^{2} x}) = b (\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}) + a (\frac{2 \times \frac{a}{b}}{1 + \frac{a^{2}}{b^{2}}}) = \frac{b (b^{2} - a^{2})}{a^{2} + b^{2}} + \frac{2 a^{2} b}{a^{2} + b^{2}}$

$= \frac{b^{3} - a^{2} b + 2 a^{2} b}{a^{2} + b^{2}} = \frac{b^{3} + a^{2} b}{a^{2} + b^{2}} = \frac{b (b^{2} + a^{2})}{a^{2} + b^{2}} = b$

Hence, the correct answer is option B.
Given: $\tan x = \frac{a}{b}$
Now,

$b \cos 2 x + a \sin 2 x = b (\frac{1 - \tan^{2} x}{1 + \tan^{2} x}) + a (\frac{2 \tan x}{1 + \tan^{2} x}) = b (\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}) + a (\frac{2 \times \frac{a}{b}}{1 + \frac{a^{2}}{b^{2}}}) = \frac{b (b^{2} - a^{2})}{a^{2} + b^{2}} + \frac{2 a^{2} b}{a^{2} + b^{2}}$

$= \frac{b^{3} - a^{2} b + 2 a^{2} b}{a^{2} + b^{2}} = \frac{b^{3} + a^{2} b}{a^{2} + b^{2}} = \frac{b (b^{2} + a^{2})}{a^{2} + b^{2}} = b$

Hence, the correct answer is option B.

Page No 9.45:

Question 37:

If $\tan α = \frac{1}{7}, \tan β = \frac{1}{3}$ , then $\cos 2 α$ is equal to

(a) $\sin 2 β$ (b) $\sin 4 β$ (c) $\sin 3 β$ (d) $\cos 2 β$

Answer:

It is given that $\tan α = \frac{1}{7}$ and $\tan β = \frac{1}{3}$ .

Now,

$\tan 2 β = \frac{2 \tan β}{1 - \tan^{2} β} = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{3}{4}$

$∴ \tan (α + 2 β) = \frac{\tan α + \tan 2 β}{1 - \tan α \tan 2 β} = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \times \frac{3}{4}} = \frac{\frac{25}{28}}{\frac{25}{28}} = 1$

$\tan (α + 2 β) = 1 = \tan \frac{π}{4} \Rightarrow α + 2 β = \frac{π}{4} \Rightarrow α = \frac{π}{4} - 2 β \Rightarrow 2 α = \frac{π}{2} - 4 β \Rightarrow \cos 2 α = \cos (\frac{π}{2} - 4 β) = \sin 4 β$

$∴ \cos 2 α = \sin 4 β$

Hence, the correct answer is option B.

Page No 9.45:

Question 38:

The value of $\cos^{2} 48 ° - \sin^{2} 12 °$ is

(a) $\frac{\sqrt{5} + 1}{8}$ (b) $\frac{\sqrt{5} - 1}{8}$ (c) $\frac{\sqrt{5} + 1}{5}$ (d) $\frac{\sqrt{5} + 1}{2 \sqrt{2}}$

Answer:

$\cos^{2} 48 ° - \sin^{2} 12 ° = \cos (48 ° + 12 °) \cos (48 ° - 12 °) [\cos (A + B) \cos (A - B) = \cos^{2} A - \sin^{2} B] = \cos 60 ° \cos 36 ° = \frac{1}{2} \times (\frac{\sqrt{5} + 1}{4}) = \frac{\sqrt{5} + 1}{8}$

Hence, the correct answer is option A.

View NCERT Solutions for all chapters of Class 15