Rs Aggarwal 2019 for Class 10 Math Chapter 10 - Trignometric Ratios

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Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 10 Trignometric Ratios are provided here with simple step-by-step explanations. These solutions for Trignometric Ratios are extremely popular among Class 10 students for Math Trignometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 539:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that sin $θ$ = $\frac{perpendicular}{hypotenuse}$ = $\frac{A B}{A C}$ = $\frac{\sqrt{3}}{2}$ .

So, if AB = $\sqrt{3} k$ , then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = (2k)² $-$ ( $\sqrt{3} k$ )²
⇒ BC² = 4k² $-$ 3k² = k²
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $θ$ = $\frac{B C}{A C}$ = $\frac{k}{2 k} = \frac{1}{2}$
tan $θ$ = $\frac{A B}{B C} = \frac{\sqrt{3} k}{k} = \sqrt{3}$

∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{1}{\sqrt{3}}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{2}{\sqrt{3}}$ and sec $θ$ = $\frac{1}{\cos θ} = 2$

Page No 539:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cos $θ$ = $\frac{Base}{hypotenuse}$ = $\frac{B C}{A C}$ = $\frac{7}{25}$ .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (25k)² $-$ (7k)².
⇒ AB² = 625k² $-$ 49k² = 576k²
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{24 k}{25 k} = \frac{24}{25}$
tan $θ$ = $\frac{A B}{B C} = \frac{24 k}{7 k} = \frac{24}{7}$
∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{7}{24}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{25}{24}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{25}{7}$

Page No 539:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{Perpendicular}{Base}$ = $\frac{A B}{B C}$ = $\frac{15}{8}$ .

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (15k)² + (8k)²
⇒ AC² = 225k² + 64k² = 289k²
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{15 k}{17 k} = \frac{15}{17}$
cos $θ$ = $\frac{B C}{A C} = \frac{8 k}{17 k} = \frac{8}{17}$

∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{8}{15}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{17}{15}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{17}{8}$

Page No 539:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cot $θ$ = $\frac{base}{Perpendicular}$ = $\frac{B C}{A B}$ = 2.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (2k)² + (k)²
⇒ AC² = 4k² + k² = 5k²
⇒ AC = $\sqrt{5}$ k
Now, finding the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{k}{\sqrt{5} k} = \frac{1}{\sqrt{5}}$
cos $θ$ = $\frac{B C}{A C} = \frac{2 k}{\sqrt{5} k} = \frac{2}{\sqrt{5}}$

∴ tan $θ$ = $\frac{1}{c o t θ} = \frac{1}{2}$ , cosec $θ$ = $\frac{1}{\sin θ} = \sqrt{5}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{\sqrt{5}}{2}$

Page No 539:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cosec $θ$ = $\frac{Hypotenuse}{Perpendicular}$ = $\frac{A C}{A B}$ = $\frac{\sqrt{10}}{1}$ .

So, if AC = ( $\sqrt{10}$ )k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = 10k² $-$ k²
⇒ BC² = 9k²
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $θ$ = $\frac{A B}{B C}$ = $\frac{k}{3 k} = \frac{1}{3}$

cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{\sqrt{10} k} = \frac{3}{\sqrt{10}}$

∴ $\sin θ = \frac{1}{\cos e c θ} = \frac{1}{\sqrt{10}}$ , cot $θ$ = $\frac{1}{\tan θ} = 3$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{\sqrt{10}}{3}$

Page No 539:

Answer:

We have $\sin θ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ ,

As,

$\cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})}^{2} = \frac{1}{1} - \frac{{(a^{2} - b^{2})}^{2}}{{(a^{2} + b^{2})}^{2}} = \frac{{(a^{2} + b^{2})}^{2} - {(a^{2} - b^{2})}^{2}}{{(a^{2} + b^{2})}^{2}} = \frac{[(a^{2} + b^{2}) - (a^{2} - b^{2})] [(a^{2} + b^{2}) + (a^{2} - b^{2})]}{{(a^{2} + b^{2})}^{2}}$
$= \frac{[a^{2} + b^{2} - a^{2} + b^{2}] [a^{2} + b^{2} + a^{2} - b^{2}]}{{(a^{2} + b^{2})}^{2}} = \frac{[2 b^{2}] [2 a^{2}]}{{(a^{2} + b^{2})}^{2}} \Rightarrow \cos^{2} θ = \frac{4 a^{2} b^{2}}{{(a^{2} + b^{2})}^{2}} \Rightarrow \cos θ = \sqrt{\frac{4 a^{2} b^{2}}{{(a^{2} + b^{2})}^{2}}} \Rightarrow \cos θ = \frac{2 a b}{(a^{2} + b^{2})}$

Also,

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

Now,

$cosec θ = \frac{1}{\sin θ} = \frac{1}{(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})} = \frac{a^{2} + b^{2}}{a^{2} - b^{2}}$

Also,

$\sec θ = \frac{1}{\cos θ} = \frac{1}{(\frac{2 a b}{a^{2} + b^{2}})} = \frac{a^{2} + b^{2}}{2 a b}$

And,

$\cot θ = \frac{1}{\tan θ} = \frac{1}{(\frac{a^{2} - b^{2}}{2 a b})} = \frac{2 a b}{a^{2} - b^{2}}$

Page No 539:

Answer:

We have,

$15 \cot A = 8 \Rightarrow \cot A = \frac{8}{15}$

As,

${cosec}^{2} A = 1 + \cot^{2} A = 1 + {(\frac{8}{15})}^{2} = 1 + \frac{64}{225} = \frac{225 + 64}{225}$
$\Rightarrow {cosec}^{2} A = \frac{289}{225} \Rightarrow cosec A = \sqrt{\frac{289}{225}} \Rightarrow cosec A = \frac{17}{15} \Rightarrow \frac{1}{\sin A} = \frac{17}{15} \Rightarrow \sin A = \frac{15}{17}$

Also,

$\cos^{2} A = 1 - \sin^{2} A = 1 - {(\frac{15}{17})}^{2} = 1 - \frac{225}{289} = \frac{289 - 225}{289}$
$\Rightarrow \cos^{2} A = \frac{64}{289} \Rightarrow \cos A = \sqrt{\frac{64}{289}} \Rightarrow \cos A = \frac{8}{17} \Rightarrow \frac{1}{\sec A} = \frac{8}{17} \Rightarrow \sec A = \frac{17}{8}$

Page No 539:

Answer:

We have $\sin A = \frac{9}{41}$ ,

As,

$\cos^{2} A = 1 - \sin^{2} A = 1 - {(\frac{9}{41})}^{2} = 1 - \frac{81}{1681} = \frac{1681 - 81}{1681}$
$\Rightarrow \cos^{2} A = \frac{1600}{1681} \Rightarrow \cos A = \sqrt{\frac{1600}{1681}} \Rightarrow \cos A = \frac{40}{41}$

Also,

$\tan A = \frac{\sin A}{\cos A} = \frac{(\frac{9}{41})}{(\frac{40}{41})} = \frac{9}{40}$

Page No 540:

Answer:

Let us consider a right $△$ ABC right angled at B.
Now, we know that cos $θ$ = 0.6 = $\frac{B C}{A C}$ = $\frac{3}{5}$

So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (5k)² $-$ (3k)² = 25k² $-$ 9k²
⇒ AB² = 16k²
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$

tan $θ$ = $\frac{A B}{B C} = \frac{4 k}{3 k} = \frac{4}{3}$

Substituting the values in the given expression, we get:
5 sin $θ$ $-$ 3 tan $θ$
$\Rightarrow 5 (\frac{4}{5}) - 3 (\frac{4}{3}) \Rightarrow 4 - 4 = 0 = RHS$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, it is given that cosec $θ$ = 2.
Also, sin $θ$ = $\frac{1}{\cos e c θ} = \frac{1}{2} = \frac{A B}{A C}$

So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC² = (2k)² $-$ (k)²
⇒ BC² = 3k²
⇒ BC = $\sqrt{3}$ k
Finding out the other T-ratios using their definitions, we get:
cos $θ$ = $\frac{B C}{A C} = \frac{\sqrt{3} k}{2 k} = \frac{\sqrt{3}}{2}$
tan $θ$ = $\frac{A B}{B C} = \frac{k}{\sqrt{3} k} = \frac{1}{\sqrt{3}}$
cot $θ$ = $\frac{1}{\tan θ} = \sqrt{3}$
Substituting these values in the given expression, we get:
$c o t θ + \frac{\sin θ}{1 + \cos θ} = \sqrt{3} + \frac{(\frac{1}{2})}{1 + \frac{\sqrt{3}}{2}} = \sqrt{3} + \frac{\frac{1}{2}}{\frac{2 + \sqrt{3}}{2}}$

$= \sqrt{3} + \frac{1}{2 + \sqrt{3}} = \frac{\sqrt{3} (2 + \sqrt{3}) + 1}{2 + \sqrt{3}} = \frac{2 \sqrt{3} + 3 + 1}{2 + \sqrt{3}} = \frac{2 (2 + \sqrt{3})}{2 + \sqrt{3}} = 2$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now it is given that tan $θ$ = $\frac{A B}{B C}$ = $\frac{1}{\sqrt{7}}$ .

So, if AB = k, then BC = $\sqrt{7}$ k, where k is a positive number.
Using Pythagoras theorem, we have:
AC²= AB² + BC²
⇒ AC² = (k)² + ( $\sqrt{7}$ k)²
⇒ AC² = k² + 7k²
⇒ AC = 2 $\sqrt{2}$ k
Now, finding out the values of the other trigonometric ratios, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{k}{2 \sqrt{2} k} = \frac{1}{2 \sqrt{2}}$
cos $θ$ = $\frac{B C}{A C} = \frac{\sqrt{7} k}{2 \sqrt{2} k} = \frac{\sqrt{7}}{2 \sqrt{2}}$
∴ cosec $θ$ = $\frac{1}{\sin θ} = 2 \sqrt{2}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{2 \sqrt{2}}{\sqrt{7}}$
Substituting the values of cosec $θ$ and sec $θ$ in the given expression, we get:
$\frac{\cos e c^{2} θ - s e c^{2} θ}{\cos e c^{2} θ + s e c^{2} θ} = \frac{(2 \sqrt{2})^{2} - {(\frac{2 \sqrt{2}}{\sqrt{7}})}^{2}}{(2 \sqrt{2})^{2} + {(\frac{2 \sqrt{2}}{\sqrt{7}})}^{2}} = \frac{8 - (\frac{8}{7})}{8 + (\frac{8}{7})} = \frac{\frac{56 - 8}{7}}{\frac{56 + 8}{7}} = \frac{48}{64} = \frac{3}{4} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (20k)²+ (21k)²
⇒ AC² = 841k²
⇒ AC = 29k
Now, sin $θ$ = $\frac{A B}{A C} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A C} = \frac{21}{29}$

Substituting these values in the given expression, we get:
$LHS = \frac{1 - \sin θ + \cos θ}{1 + \sin θ + \cos θ} = \frac{1 - \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29 - 20 + 21}{29}}{\frac{29 + 20 + 21}{29}} = \frac{30}{70} = \frac{3}{7} = RHS$
∴ LHS = RHS

Hence proved.

Page No 540:

Answer:

We have,

$\sec θ = \frac{5}{4} \Rightarrow \frac{1}{\cos θ} = \frac{5}{4} \Rightarrow \cos θ = \frac{4}{5}$

Also,

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

Now,

$LHS = \frac{(\sin θ - 2 \cos θ)}{(\tan θ - \cot θ)} = \frac{(\sin θ - 2 \cos θ)}{(\frac{\sin θ}{\cos θ} - \frac{\cos θ}{\sin θ})} = \frac{(\sin θ - 2 \cos θ)}{(\frac{\sin^{2} θ - \cos^{2} θ}{\sin θ \cos θ})} = \frac{\sin θ \cos θ (\sin θ - 2 \cos θ)}{(\sin^{2} θ - \cos^{2} θ)}$
$= \frac{\frac{3}{5} \times \frac{4}{5} (\frac{3}{5} - 2 \times \frac{4}{5})}{{(\frac{3}{5})}^{2} - {(\frac{4}{5})}^{2}} = \frac{\frac{12}{25} (\frac{3}{5} - \frac{8}{5})}{(\frac{9}{25} - \frac{16}{25})} = \frac{\frac{12}{25} \times (\frac{- 5}{5})}{(\frac{- 7}{25})} = \frac{12}{7} = RHS$

Page No 540:

Answer:

$LHS = \sqrt{\frac{\sec θ - cosec θ}{\sec θ + cosec θ}} = \sqrt{\frac{(\frac{1}{\cos θ} - \frac{1}{\sin θ})}{(\frac{1}{\cos θ} + \frac{1}{\sin θ})}} = \sqrt{\frac{(\frac{\sin θ - \cos θ}{\sin θ \cos θ})}{(\frac{\sin θ + \cos θ}{\sin θ \cos θ})}} = \sqrt{\frac{(\frac{\sin θ - \cos θ}{\sin θ})}{(\frac{\sin θ + \cos θ}{\sin θ})}} = \sqrt{\frac{(\frac{\sin θ}{\sin θ} - \frac{\cos θ}{\sin θ})}{(\frac{\sin θ}{\sin θ} + \frac{\cos θ}{\sin θ})}} = \sqrt{\frac{1 - \cot θ}{1 + \cot θ}} = \sqrt{\frac{(1 - \frac{3}{4})}{(1 + \frac{3}{4})}}$
$= \sqrt{\frac{(\frac{1}{4})}{(\frac{7}{4})}} = \sqrt{\frac{1}{7}} = \frac{1}{\sqrt{7}} = RHS$

Page No 540:

Answer:

$LHS = \sqrt{\frac{{cosec}^{2} θ - \cot^{2} θ}{\sec^{2} θ - 1}} = \sqrt{\frac{1}{\tan^{2} θ}} = \sqrt{\cot^{2} θ} = \cot θ = \sqrt{{cosec}^{2} θ - 1} = \sqrt{{(\frac{1}{\sin θ})}^{2} - 1} = \sqrt{{(\frac{1}{(\frac{3}{4})})}^{2} - 1} = \sqrt{{(\frac{4}{3})}^{2} - 1} = \sqrt{\frac{16}{9} - 1} = \sqrt{\frac{16 - 9}{9}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3} = RHS$

Page No 540:

Answer:

$LHS = (\sec θ + \tan θ) = \frac{1}{\cos θ} + \frac{\sin θ}{\cos θ} = \frac{1 + \sin θ}{\cos θ} = \frac{1 + \sin θ}{\sqrt{1 - \sin^{2} θ}} = \frac{(1 + \frac{a}{b})}{\sqrt{1 - {(\frac{a}{b})}^{2}}}$
$= \frac{(\frac{1}{1} + \frac{a}{b})}{\sqrt{\frac{1}{1} - \frac{a^{2}}{b^{2}}}} = \frac{(\frac{b + a}{b})}{\sqrt{\frac{b^{2} - a^{2}}{b^{2}}}} = \frac{(\frac{b + a}{b})}{(\frac{\sqrt{b^{2} - a^{2}}}{b})} = \frac{(b + a)}{\sqrt{(b + a) (b - a)}}$
$= \frac{(b + a)}{\sqrt{(b + a)} \sqrt{(b - a)}} = \frac{\sqrt{(b + a)}}{\sqrt{(b - a)}} = \sqrt{\frac{b + a}{b - a}} = RHS$

Page No 540:

Answer:

$LHS = \frac{(\sin θ - \cot θ)}{2 \tan θ} = \frac{(\sin θ - \frac{\cos θ}{\sin θ})}{2 (\frac{\sin θ}{\cos θ})} = \frac{(\frac{\sin^{2} θ - \cos θ}{\sin θ})}{(\frac{2 \sin θ}{\cos θ})} = \frac{\cos θ (\sin^{2} θ - \cos θ)}{2 \sin^{2} θ} = \frac{\cos θ (1 - \cos^{2} θ - \cos θ)}{2 (1 - \cos^{2} θ)}$
$= \frac{\frac{3}{5} [1 - {(\frac{3}{5})}^{2} - \frac{3}{5}]}{2 [1 - {(\frac{3}{5})}^{2}]} = \frac{\frac{3}{5} (\frac{1}{1} - \frac{9}{25} - \frac{3}{5})}{2 (1 - \frac{9}{25})} = \frac{\frac{3}{5} (\frac{25 - 9 - 15}{25})}{2 (\frac{25 - 9}{25})} = \frac{\frac{3}{5} (\frac{1}{25})}{2 (\frac{16}{25})}$
$= \frac{3}{5 \times 2 \times 16} = \frac{3}{160} = RHS$

Page No 540:

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{4}{3}$ .

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC² = (4k)² + (3k)²
⇒ AC² = 16k²+ 9k² = 25k²
⇒ AC = 5k
Finding out the values of sin $θ$ and cos $θ$ using their definitions, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$
cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$
Substituting these values in the given expression, we get:
(sin $θ$ + cos $θ$ ) = $(\frac{4}{5} + \frac{3}{5}) = (\frac{7}{5}) = RHS$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

It is given that tan $θ = \frac{a}{b}$ .

LHS = $\frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ}$
Dividing the numerator and denominator by cos $θ$ , we get:

$\frac{a \tan θ - b}{a \tan θ + b}$ (∵ tan $θ = \frac{\sin θ}{\cos θ}$ )
Now, substituting the value of tan $θ$ in the above expression, we get:
$\frac{a (\frac{a}{b}) - b}{a (\frac{a}{b}) + b} = \frac{\frac{a^{2}}{b} - b}{\frac{a^{2}}{b} + b} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
We know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{4}{3}$

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AC² = 16k² + 9k²
⇒ AC² = 25k²
⇒ AC = 5k

Now, we have:

sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$

cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$

Substituting these values in the given expression, we get:

$\frac{4 \cos θ - \sin θ}{2 \cos θ + \sin θ} = \frac{4 (\frac{3}{5}) - \frac{4}{5}}{2 (\frac{3}{5}) + \frac{4}{5}} = \frac{\frac{12}{5} - \frac{4}{5}}{\frac{6}{5} + \frac{4}{5}} = \frac{\frac{12 - 4}{5}}{\frac{6 + 4}{5}} = \frac{8}{10} = \frac{4}{5} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

It is given that cot $θ = \frac{2}{3}$ .

LHS = $\frac{4 \sin θ - 3 \cos θ}{2 \sin θ + 6 \cos θ}$
Dividing the above expression by sin $θ$ , we get:
$\frac{4 - 3 c o t θ}{2 + 6 c o t θ}$ [∵ cot $θ = \frac{\cos θ}{\sin θ}$ ]
Now, substituting the values of cot $θ$ in the above expression, we get:
$\frac{4 - 3 (\frac{2}{3})}{2 + 6 (\frac{2}{3})} = \frac{4 - 2}{2 + 4} = \frac{2}{6} = \frac{1}{3}$
i.e., LHS = RHS

Hence proved.

Page No 540:

Answer:

$LHS = \frac{(1 - \tan^{2} θ)}{(1 + \tan^{2} θ)} = \frac{(1 - \frac{1}{\cot^{2} θ})}{(1 + \frac{1}{\cot^{2} θ})} = \frac{(\frac{\cot^{2} θ - 1}{\cot^{2} θ})}{(\frac{\cot^{2} θ + 1}{\cot^{2} θ})} = \frac{\cot^{2} θ - 1}{\cot^{2} θ + 1} = \frac{{(\frac{4}{3})}^{2} - 1}{{(\frac{4}{3})}^{2} + 1} (As, 3 \cot θ = 4 or \cot θ = \frac{4}{3})$
$= \frac{\frac{16}{9} - 1}{\frac{16}{9} + 1} = \frac{(\frac{16 - 9}{9})}{(\frac{16 + 9}{9})} = \frac{(\frac{7}{9})}{(\frac{25}{9})} = \frac{7}{25}$
$RHS = (\cos^{2} θ - \sin^{2} θ) = \frac{(\cos^{2} θ - \sin^{2} θ)}{1} = \frac{(\frac{\cos^{2} θ - \sin^{2} θ}{\sin^{2} θ})}{(\frac{1}{\sin^{2} θ})} = \frac{(\frac{\cos^{2} θ}{\sin^{2} θ} - \frac{\sin^{2} θ}{\sin^{2} θ})}{{cosec}^{2} θ} = \frac{(\cot^{2} θ - 1)}{(\cot^{2} θ + 1)}$
$= \frac{[{(\frac{4}{3})}^{2} - 1]}{[{(\frac{4}{3})}^{2} + 1]} = \frac{(\frac{16}{9} - \frac{1}{1})}{(\frac{16}{9} + \frac{1}{1})} = \frac{(\frac{16 - 9}{9})}{(\frac{16 + 9}{9})} = \frac{(\frac{7}{9})}{(\frac{25}{9})} = \frac{7}{25}$

Since, LHS=RHS
Hence, verified.

Page No 540:

Answer:

It is given that sec $θ$ = $\frac{17}{8}$ .

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
We know that cos $θ$ = $\frac{1}{s e c θ} = \frac{8}{17} = \frac{B C}{A C}$

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (17k)² $-$ (8k)²
⇒ AB² = 289k² $-$ 64k² = 225k²
⇒ AB = 15k.

Now, tan $θ$ = $\frac{A B}{B C} = \frac{15}{8}$ and sin $θ$ = $\frac{A B}{A C} = \frac{15 k}{17 k} = \frac{15}{17}$

The given expression is $\frac{3 - 4 \sin^{2} θ}{4 \cos^{2} θ - 3} = \frac{3 - \tan^{2} θ}{1 - 3 \tan^{2} θ}$ .

Substituting the values in the above expression, we get:
$LHS = \frac{3 - 4 {(\frac{15}{17})}^{2}}{4 {(\frac{8}{17})}^{2} - 3} = \frac{3 - \frac{900}{289}}{\frac{256}{289} - 3} = \frac{867 - 900}{256 - 867} = \frac{- 33}{- 611} = \frac{33}{611}$

$RHS = \frac{3 - {(\frac{15}{8})}^{2}}{1 - 3 {(\frac{15}{8})}^{2}} = \frac{3 - \frac{225}{64}}{1 - \frac{675}{64}} = \frac{192 - 225}{64 - 675} = \frac{- 33}{- 611} = \frac{33}{611}$

∴ LHS = RHS
Hence proved.

Page No 540:

Answer:

In $∆$ ABD,

Using Pythagoras theorem, we get

$AB = \sqrt{{AD}^{2} - {BD}^{2}} = \sqrt{10^{2} - 8^{2}} = \sqrt{100 - 64} = \sqrt{36} = 6 cm$

Again,

In $∆$ ABC,

Using Pythagoras therem, we get

$AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{6^{2} + 4^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13} cm$

Now,

$(i) \sin θ = \frac{BC}{AC} = \frac{4}{2 \sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2 \sqrt{13}}{13}$

$(ii) \cos θ = \frac{AB}{AC} = \frac{6}{2 \sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3 \sqrt{13}}{13}$

Page No 540:

Answer:

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (24)² + (7)²
⇒ AC² = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) sin A = $\frac{B C}{A C} = \frac{7}{25}$

(ii) cos A = $\frac{A B}{A C} = \frac{24}{25}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C = $\frac{A B}{A C} = \frac{24}{25}$

(iv) cos C = $\frac{B C}{A C} = \frac{7}{25}$

Page No 541:

Answer:

Using Pythagoras theorem, we get:
AB²=AC²+ BC²
⇒ AC² = AB² $-$ BC²
⇒ AC² = (29)² $-$ (21)²
⇒ AC² = 841 $-$ 441
⇒ AC²= 400
⇒ AC = $\sqrt{400}$ = 20 units

Now, sin $θ = \frac{A C}{A B} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A B} = \frac{21}{29}$

cos² $θ$ $-$ sin² $θ$ = ${(\frac{21}{29})}^{2} - {(\frac{20}{29})}^{2} = \frac{441}{841} - \frac{400}{841} = \frac{41}{841}$

Hence Proved.

Page No 541:

Answer:

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = 12² + 5² = 144 + 25
⇒ AC² = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, AB is base and BC is perpendicular
(i) cosA = $\frac{AB}{AC} = \frac{12}{13}$
(ii) cosecA = $\frac{1}{sinA} = \frac{AC}{BC} = \frac{13}{5}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cosC = $\frac{BC}{AC} = \frac{5}{13}$
(iv) cosecC = $\frac{1}{sinC} = \frac{AC}{AB} = \frac{13}{12}$

Page No 541:

Answer:

$LHS = (3 \cos α - 4 \cos^{3} α) = \cos α (3 - 4 \cos^{2} α) = \sqrt{1 - \sin^{2} α} [3 - 4 (1 - \sin^{2} α)] = \sqrt{1 - {(\frac{1}{2})}^{2}} [3 - 4 (1 - {(\frac{1}{2})}^{2})] = \sqrt{\frac{1}{1} - \frac{1}{4}} [3 - 4 (\frac{1}{1} - \frac{1}{4})] = \sqrt{\frac{3}{4}} [3 - 4 (\frac{3}{4})] = \sqrt{\frac{3}{4}} [3 - 3] = \sqrt{\frac{3}{4}} [0] = 0 = RHS$

Page No 541:

Answer:

$In ∆ ABC, ∠ B = 90 °, As, tanA = \frac{1}{\sqrt{3}} \Rightarrow \frac{BC}{AB} = \frac{1}{\sqrt{3}} Let BC = x and AB = x \sqrt{3} Using Pythagoras theorem, we get AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{{(x \sqrt{3})}^{2} + x^{2}} = \sqrt{3 x^{2} + x^{2}} = \sqrt{4 x^{2}} = 2 x$

Now,

$(i) LHS = sinA \cdot cosC + cosA \cdot sinC = \frac{BC}{AC} \cdot \frac{BC}{AC} + \frac{AB}{AC} \cdot \frac{AB}{AC} = {(\frac{BC}{AC})}^{2} + {(\frac{AB}{AC})}^{2} = {(\frac{x}{2 x})}^{2} + {(\frac{x \sqrt{3}}{2 x})}^{2} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 = RHS$

$(ii) LHS = cosA \cdot cosC - sinA \cdot sinC = \frac{AB}{AC} \cdot \frac{BC}{AC} - \frac{BC}{AC} \cdot \frac{AB}{AC} = \frac{x \sqrt{3}}{2 x} . \frac{x}{2 x} - \frac{x}{2 x} . \frac{x \sqrt{3}}{2 x} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 = RHS$

Page No 541:

Answer:

In $∆$ ABC, $∠$ C = 90 $°$
sinA = $\frac{BC}{AB}$ and
sinB = $\frac{AC}{AB}$

As, sinA = sinB
$\Rightarrow$ $\frac{BC}{AB}$ = $\frac{AC}{AB}$
$\Rightarrow$ BC = AC
So, $∠$ A = $∠$ B (Angles opposite to equal sides are equal)

Page No 541:

Answer:

In $∆ ABC, ∠ C = 90 °$ ,

$tanA = \frac{BC}{AC} and tanB = \frac{AC}{BC}$

$As, tanA = tanB$
$\Rightarrow \frac{BC}{AC} = \frac{AC}{BC} \Rightarrow {BC}^{2} = {AC}^{2} \Rightarrow BC = AC So, ∠ A = ∠ B (Angles opposite to equal sides are equal)$

Page No 541:

Answer:

$We have, tanA = 1 \Rightarrow \frac{sinA}{cosA} = 1 \Rightarrow sinA = cosA \Rightarrow sinA - cosA = 0 Squaring both sides, we get {(sinA - cosA)}^{2} = 0 \Rightarrow \sin^{2} A + \cos^{2} A - 2 sinA \cdot cosA = 0 \Rightarrow 1 - 2 sinA \cdot cosA = 0 ∴ 2 sinA \cdot cosA = 1$

Page No 541:

Answer:

In $∆ PQR, ∠ Q = 90 °$ ,

Using Pythagoras theorem, we get

$PQ = \sqrt{{PR}^{2} - {QR}^{2}} = \sqrt{{(x + 2)}^{2} - x^{2}} = \sqrt{x^{2} + 4 x + 4 - x^{2}} = \sqrt{4 (x + 1)} = 2 \sqrt{x + 1}$

Now,

$(i) (\sqrt{x + 1}) \cot ϕ = (\sqrt{x + 1}) \times \frac{QR}{PQ} = (\sqrt{x + 1}) \times \frac{x}{2 \sqrt{x + 1}} = \frac{x}{2}$

$(ii) (\sqrt{x^{3} + x^{2}}) \tan θ = (\sqrt{x^{2} (x + 1)}) \times \frac{QR}{PQ} = x \sqrt{(x + 1)} \times \frac{x}{2 \sqrt{x + 1}} = \frac{x^{2}}{2}$

$(iii) \cos θ = \frac{PQ}{PR} = \frac{2 \sqrt{x + 1}}{(x + 2)}$

Page No 541:

Answer:

$LHS = {(\frac{2}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} - 1 = {[\frac{2}{(cosecA + cosA) + (cosecA - cosA)}]}^{2} + {[\frac{(cosecA + cosA) - (cosecA - cosA)}{2}]}^{2} - 1 = {[\frac{2}{cosecA + cosA + cosecA - cosA}]}^{2} + {[\frac{cosecA + cosA - cosecA + cosA}{2}]}^{2} - 1 = {[\frac{2}{2 cosecA}]}^{2} + {[\frac{2 cosA}{2}]}^{2} - 1$
$= {[\frac{1}{cosecA}]}^{2} + {[cosA]}^{2} - 1 = {[sinA]}^{2} + {[cosA]}^{2} - 1 = \sin^{2} A + \cos^{2} A - 1 = 1 - 1 = 0 = RHS$

Page No 541:

Answer:

$LHS = {(\frac{x - y}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} = {[\frac{(cotA + cosA) - (cotA - cosA)}{(cotA + cosA) + (cotA - cosA)}]}^{2} + {[\frac{(cotA + cosA) - (cotA - cosA)}{2}]}^{2} = {[\frac{cotA + cosA - cotA + cosA}{cotA + cosA + cotA - cosA}]}^{2} + {[\frac{cotA + cosA - cotA + cosA}{2}]}^{2} = {[\frac{2 cosA}{2 cotA}]}^{2} + {[\frac{2 cosA}{2}]}^{2} = {[\frac{cosA}{(\frac{cosA}{sinA})}]}^{2} + {[cosA]}^{2} = {[\frac{sinA cosA}{cosA}]}^{2} + {[cosA]}^{2} = {[sinA]}^{2} + {[cosA]}^{2} = \sin^{2} A + \cos^{2} A = 1 = RHS$

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