Rs Aggarwal 2020 2021 for Class 10 Maths Chapter 13 - Trigonometric Identities

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Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 13 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Maths Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 617:

Answer:

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

Page No 617:

Answer:

We know
$\sin^{2} θ + \cos^{2} θ = 1$
Squaring on both sides, we get
${(\sin^{2} θ + \cos^{2} θ)}^{2} = 1 \Rightarrow {(\sin^{2} θ)}^{2} + {(\cos^{2} θ)}^{2} + 2 \sin^{2} θ \cos^{2} θ = 1 [{(a + b)}^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow \sin^{4} θ + \cos^{4} θ = 1 - 2 \sin^{2} θ \cos^{2} θ$
$∴ \frac{\sin^{4} θ + \cos^{4} θ}{1 - 2 \sin^{2} θ \cos^{2} θ} = \frac{1 - 2 \sin^{2} θ \cos^{2} θ}{1 - 2 \sin^{2} θ \cos^{2} θ} = 1$

Page No 617:

Answer:

Ans

Page No 617:

Answer:

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

Page No 617:

Answer:

$\frac{1 - \sin θ}{1 + \sin θ} = \frac{1 - \sin θ}{1 + \sin θ} \times \frac{1 - \sin θ}{1 - \sin θ} = \frac{{(1 - \sin θ)}^{2}}{1 - \sin^{2} θ} [(a + b) (a - b) = a^{2} - b^{2}]$
$= \frac{{(1 - \sin θ)}^{2}}{\cos^{2} θ} (\sin^{2} θ + \cos^{2} θ = 1) = {(\frac{1 - \sin θ}{\cos θ})}^{2} = {(\frac{1}{\cos θ} - \frac{\sin θ}{\cos θ})}^{2} = {(\sec θ - \tan θ)}^{2}$

Page No 617:

Answer:

Ans

Page No 617:

Answer:

$\sec^{4} θ - \tan^{4} θ = {(\sec^{2} θ)}^{2} - {(\tan^{2} θ)}^{2} = (\sec^{2} θ - \tan^{2} θ) (\sec^{2} θ + \tan^{2} θ) [a^{2} - b^{2} = (a - b) (a + b)]$
$= 1 \times (1 + \tan^{2} θ + \tan^{2} θ) (1 + \tan^{2} θ = \sec^{2} θ) = 1 + 2 \tan^{2} θ$

Page No 617:

Answer:

$\begin{array}{l} LHS= \frac{sin θ}{(cot θ +cosec θ)} - \frac{sin θ}{(cot θ - cosec θ)} \\ =sin θ {\frac{(cot θ - cosec θ) - (cot θ + cosec θ)}{(cot θ +cosec θ) (cot θ - cosec θ)}} \\ =sin θ {\frac{- 2 cosec θ}{({cot}^{2} θ - {cosec}^{2} θ)}} \\ =sin θ (\frac{- 2 cosec θ}{- 1}) (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ =sin θ .2cosec θ \\ =sin θ ×2× \frac{1}{sin θ} \\ =2 \\ =RHS \end{array}$

Page No 617:

Answer:

$\frac{cosec A - \sin A}{cosec A + \sin A} = \frac{\frac{1}{\sin A} - \sin A}{\frac{1}{\sin A} + \sin A} = \frac{\frac{1 - \sin^{2} A}{\sin A}}{\frac{1 + \sin^{2} A}{\sin A}} = \frac{1 - \sin^{2} A}{1 + \sin^{2} A}$
$= \frac{\frac{1 - \sin^{2} A}{\cos^{2} A}}{\frac{1 + \sin^{2} A}{\cos^{2} A}}$ (Dividing numerator and denominator by cos²A)
$= \frac{\frac{1}{\cos^{2} A} - \frac{\sin^{2} A}{\cos^{2} A}}{\frac{1}{\cos^{2} A} + \frac{\sin^{2} A}{\cos^{2} A}} = \frac{\sec^{2} A - \tan^{2} A}{\sec^{2} A + \tan^{2} A} (\sec θ = \frac{1}{\cos θ} and \tan θ = \frac{\sin θ}{\cos θ})$

Page No 617:

Answer:

$\sin^{2} θ \tan θ + \cos^{2} θ \cot θ + 2 \sin θ \cos θ = \frac{\sin^{3} θ}{\cos θ} + \frac{\cos^{3} θ}{\sin θ} + 2 \sin θ \cos θ = \frac{\sin^{4} θ + \cos^{4} θ + 2 \sin^{2} θ \cos^{2} θ}{\sin θ \cos θ} = \frac{{(\sin^{2} θ + \cos^{2} θ)}^{2}}{\sin θ \cos θ}$
$= \frac{1}{\sin θ \cos θ} (\sin^{2} θ + \cos^{2} θ = 1) = \frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ} = \frac{\sin^{2} θ}{\sin θ \cos θ} + \frac{\cos^{2} θ}{\sin θ \cos θ}$
$= \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ} = \tan θ + \cot θ$

Page No 617:

Answer:

$(\tan θ + \sec θ - 1) (\tan θ + \sec θ + 1) = {(\tan θ + \sec θ)}^{2} - 1 [(a - b) (a + b) = a^{2} - b^{2}] = \tan^{2} θ + \sec^{2} θ + 2 \tan θ \sec θ - 1 = 2 \tan^{2} θ + 2 \tan θ \sec θ (1 + \tan^{2} θ = \sec^{2} θ)$
$= 2 \tan θ (\tan θ + \sec θ) = 2 \times \frac{\sin θ}{\cos θ} \times (\frac{\sin θ}{\cos θ} + \frac{1}{\cos θ}) = \frac{2 \sin θ (1 + \sin θ)}{\cos^{2} θ} = \frac{2 \sin θ (1 + \sin θ)}{1 - \sin^{2} θ} (\sin^{2} θ + \cos^{2} θ = 1)$
$= \frac{2 \sin θ (1 + \sin θ)}{(1 - \sin θ) (1 + \sin θ)} = \frac{2 \sin θ}{1 - \sin θ}$

Page No 617:

Answer:

$(1 + \cot A + \tan A) (\sin A - \cos A) = \sin A + \cot A \sin A + \tan A \sin A - \cos A - \cot A \cos A - \tan A \cos A = \sin A + \frac{\cos A}{\sin A} \times \sin A + \tan A \sin A - \cos A - \cot A \cos A - \frac{\sin A}{\cos A} \times \cos A$
$= \sin A + \cos A + \tan A \sin A - \cos A - \cot A \cos A - \sin A = \sin A \tan A - \cot A \cos A$

Page No 617:

Answer:

Ans

Page No 617:

Answer:

$\sec^{2} θ - \frac{\sin^{2} θ - 2 \sin^{4} θ}{2 \cos^{4} θ - \cos^{2} θ} = \sec^{2} θ - \frac{\sin^{2} θ (1 - 2 \sin^{2} θ)}{\cos^{2} θ (2 \cos^{2} θ - 1)} = \sec^{2} θ - \frac{\sin^{2} θ (1 - 2 \sin^{2} θ)}{\cos^{2} θ [2 (1 - \sin^{2} θ) - 1]} [\cos^{2} θ + \sin^{2} θ = 1]$
$= \sec^{2} θ - \frac{\sin^{2} θ (1 - 2 \sin^{2} θ)}{\cos^{2} θ (1 - 2 \sin^{2} θ)} = \sec^{2} θ - \tan^{2} θ = 1 (1 + \tan^{2} θ = \sec^{2} θ)$

Page No 617:

Answer:

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Page No 617:

Answer:

$\frac{\tan A + \sin A}{\tan A - \sin A} = \frac{\frac{\sin A}{\cos A} + \sin A}{\frac{\sin A}{\cos A} - \sin A} = \frac{\sin A (\frac{1}{\cos A} + 1)}{\sin A (\frac{1}{\cos A} - 1)} = \frac{\sec A + 1}{\sec A - 1} (\sec A = \frac{1}{\cos A})$

Page No 617:

Answer:

$\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A} = \frac{\cos A (\frac{1}{\sin A} - 1)}{\cos A (\frac{1}{\sin A} + 1)} = \frac{cosec A - 1}{cosec A + 1} (cosec A = \frac{1}{\sin A})$

Page No 617:

Answer:

$\frac{1 - \sin θ}{1 + \sin θ} = \frac{1 - \sin θ}{1 + \sin θ} \times \frac{1 - \sin θ}{1 - \sin θ} = \frac{{(1 - \sin θ)}^{2}}{(1 + \sin θ) (1 - \sin θ)} = \frac{{(1 - \sin θ)}^{2}}{1 - \sin^{2} θ} [(a + b) (a - b) = a^{2} - b^{2}]$
$= \frac{{(1 - \sin θ)}^{2}}{\cos^{2} θ} (\sin^{2} θ + \cos^{2} θ = 1) = {(\frac{1 - \sin θ}{\cos θ})}^{2} = {(\frac{1}{\cos θ} - \frac{\sin θ}{\cos θ})}^{2} = {(\sec θ - \tan θ)}^{2}$

Page No 617:

Answer:

$\frac{1 + \cos θ}{1 - \cos θ} = \frac{1 + \cos θ}{1 - \cos θ} \times \frac{1 + \cos θ}{1 + \cos θ} = \frac{{(1 + \cos θ)}^{2}}{(1 - \cos θ) (1 + \cos θ)} = \frac{{(1 + \cos θ)}^{2}}{1 - \cos^{2} θ} [(a + b) (a - b) = a^{2} - b^{2}]$
$= \frac{{(1 + \cos θ)}^{2}}{\sin^{2} θ} (\sin^{2} θ + \cos^{2} θ = 1) = {(\frac{1 + \cos θ}{\sin θ})}^{2} = {(\frac{1}{\sin θ} + \frac{\cos θ}{\sin θ})}^{2} = {(cosec θ + \cot θ)}^{2}$

Page No 617:

Answer:

$\frac{\sec θ - \tan θ}{\sec θ + \tan θ} = \frac{\sec θ - \tan θ}{\sec θ + \tan θ} \times \frac{\sec θ - \tan θ}{\sec θ - \tan θ} = \frac{{(\sec θ - \tan θ)}^{2}}{(\sec θ + \tan θ) (\sec θ - \tan θ)} = \frac{\sec^{2} θ + \tan^{2} θ - 2 \sec θ \tan θ}{\sec^{2} θ - \tan^{2} θ} [{(a - b)}^{2} = a^{2} + b^{2} - 2 a b and (a + b) (a - b) = a^{2} - b^{2}]$
$= \frac{1 + \tan^{2} θ + \tan^{2} θ - 2 \sec θ \tan θ}{1} (1 + \tan^{2} θ = \sec^{2} θ) = 1 + 2 \tan^{2} θ - 2 \sec θ \tan θ$

Page No 618:

Answer:

ANS

Page No 618:

Answer:

$\begin{array}{l} LHS= \frac{{cos}^{3} θ {+sin}^{3} θ}{cos θ +sin θ} + \frac{{cos}^{3} θ - {sin}^{3} θ}{cos θ - sin θ} \\ = \frac{(cos θ +sin θ) ({cos}^{2} θ - cos θ sin θ {+sin}^{2} θ)}{(cos θ +sin θ)} + \frac{(cos θ - sin θ) ({cos}^{2} θ + cos θ sin θ {+sin}^{2} θ)}{(cos θ - sin θ)} \\ = ({cos}^{2} θ {+sin}^{2} θ - cos θ sin θ) + ({cos}^{2} θ {+sin}^{2} θ + cos θ sin θ) \\ = (1 - cos θ sin θ) + (1 + cos θ sin θ) \\ =2 \\ =RHS \end{array}$
Hence, LHS= RHS

Page No 618:

Answer:

$\begin{array}{l} LHS= \frac{1 + cos θ - {sin}^{2} θ}{sin θ (1 + cos θ)} \\ = \frac{(1 + cos θ) - (1 - {cos}^{2} θ)}{sin θ (1 + cos θ)} \\ = \frac{cos θ {+cos}^{2} θ}{sin θ (1 + cos θ)} \\ = \frac{cos θ (1 + cos θ)}{sin θ (1 + cos θ)} \\ = \frac{cos θ}{sin θ} \\ =cot θ \\ =RHS \end{array}$
Hence, L.H.S. = R.H.S.

Page No 618:

Answer:

ANS

Page No 618:

Answer:

ANS

Page No 618:

Answer:

Ans

Page No 618:

Answer:

Ans

Page No 618:

Answer:

$\begin{array}{l} LHS= \frac{cos θ cosec θ - sin θ sec θ}{cos θ +sin θ} \\ = \frac{\frac{cos θ}{sin θ} - \frac{sin θ}{cos θ}}{cos θ +sin θ} \\ = \frac{{cos}^{2} θ - {sin}^{2} θ}{cos θ sin θ (cos θ +sin θ)} \\ = \frac{(cos θ +sin θ) (cos θ - sin θ)}{cos θ sin θ (cos θ +sin θ)} \\ = \frac{(cos θ - sin θ)}{cos θ sin θ} \\ = \frac{1}{sin θ} - \frac{1}{cos θ} \\ =cosec θ - sec θ \\ =RHS \end{array}$
Hence, LHS = RHS

Page No 618:

Answer:

$\begin{array}{l} LHS= (1 + tan θ +cot θ) (sin θ - cos θ) \\ =sin θ +tan θ sin θ + cot θ sin θ - cos θ - tan θ cos θ - cot θ cos θ \\ =sin θ +tan θ sin θ + \frac{cos θ}{sin θ} \times sin θ - cos θ - \frac{\sin θ}{cos θ} \times cos θ - cot θ cos θ \\ =sin θ +tan θ sin θ +cos θ - cos θ - sin θ - cot θ cos θ \\ =tan θ sin θ - cot θ cos θ \\ = \frac{sin θ}{cos θ} \times \frac{1}{cosec θ} - \frac{cos θ}{sin θ} \times \frac{1}{sec θ} \\ = \frac{1}{cosec θ} \times \frac{1}{cosec θ} \times sec θ - \frac{1}{sec θ} \times \frac{1}{sec θ} \times cosec θ \\ = \frac{sec θ}{{cosec}^{2} θ} - \frac{cosec θ}{{sec}^{2} θ} \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 618:

Answer:

$LHS = \frac{\cot^{2} θ (\sec θ - 1)}{(1 + \sin θ)} + \frac{\sec^{2} θ (\sin θ - 1)}{(1 + \sec θ)} = \frac{\frac{\cos^{2} θ}{\sin^{2} θ} (\frac{1}{\cos θ} - 1)}{(1 + \sin θ)} + \frac{\frac{1}{\cos^{2} θ} (\sin θ - 1)}{(1 + \frac{1}{\cos θ})} = \frac{\frac{\cos^{2} θ}{\sin^{2} θ} (\frac{1 - \cos θ}{\cos θ})}{(1 + \sin θ)} + \frac{\frac{(\sin θ - 1)}{\cos^{2} θ}}{(\frac{\cos θ + 1}{\cos θ})} = \frac{\cos^{2} θ (1 - \cos θ)}{\sin^{2} θ \cos θ (1 + \sin θ)} + \frac{(\sin θ - 1) \cos θ}{(\cos θ + 1) \cos^{2} θ} = \frac{\cos θ (1 - \cos θ)}{(1 - \cos^{2} θ) (1 + \sin θ)} + \frac{(\sin θ - 1) \cos θ}{(\cos θ + 1) (1 - \sin^{2} θ)} = \frac{\cos θ (1 - \cos θ)}{(1 - \cos θ) (1 + \cos θ) (1 + \sin θ)} + \frac{- (1 - \sin θ) \cos θ}{(\cos θ + 1) (1 - \sin θ) (1 + \sin θ)} = \frac{\cos θ}{(1 + \cos θ) (1 + \sin θ)} - \frac{\cos θ}{(\cos θ + 1) (1 + \sin θ)} = 0 = RHS$

Page No 618:

Answer:

$\begin{array}{l} (i) {cos}^{2} θ + cos θ =1 \\ {LHS=cos}^{2} θ + cos θ \\ =1 - {sin}^{2} θ + cos θ \\ =1 - ({sin}^{2} θ - cos θ) \\ Since LHS \neq RHS, this is not an identity. \\ (ii) {sin}^{2} θ + sin θ =1 \\ {LHS=sin}^{2} θ + sin θ \\ =1 - {cos}^{2} θ + sin θ \\ =1 - ({cos}^{2} θ - sin θ) \\ Since LHS≠RHS, this is not an identity . \\ (iii) {tan}^{2} θ + sin θ {=cos}^{2} θ \\ {LHS=tan}^{2} θ + sin θ \\ = \frac{{sin}^{2} θ}{{cos}^{2} θ} + sin θ \\ = \frac{1 - {cos}^{2} θ}{{cos}^{2} θ} + sin θ \\ {=sec}^{2} θ - 1 + sin θ \\ Since LHS \neq RHS, this is not an identity . \end{array}$

Page No 628:

Answer:

$\begin{array}{l} We have m^{2} + n^{2} = [{(a \cos θ + b \sin θ)}^{2} + {(a \sin θ - b \cos θ)}^{2}] \\ = (a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + 2 a b \cos θ \sin θ) + (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ - 2 a b \sin θ \cos θ) \\ = a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + a^{2} \sin^{2} θ + b^{2} \cos^{2} θ \\ =(a^{2} \cos^{2} θ + a^{2} \sin^{2} θ) + (b^{2} \cos^{2} θ + b^{2} \sin^{2} θ) \\ = a^{2} (\cos^{2} θ + \sin^{2} θ) + b^{2} (\cos^{2} θ + \sin^{2} θ) \\ = a^{2} + b^{2} [∵ {sin}^{2} + \cos^{2} = 1] \\ {Hence, m}^{2} + n^{2} = a^{2} + b^{2} \end{array}$

Page No 628:

Answer:

$\begin{array}{l} We have x^{2} - y^{2} = [{(a s e c θ + b \tan θ)}^{2} - {(a \tan θ + b s e c θ)}^{2}] \\ = (a^{2} \sec^{2} θ + b^{2} \tan^{2} θ + 2 a b \sec θ \tan θ) - (a^{2} \tan^{2} θ + b^{2} \sec^{2} θ + 2 a b \tan θ \sec θ) \\ = a^{2} \sec^{2} θ + b^{2} \tan^{2} θ - a^{2} \tan^{2} θ - b^{2} \sec^{2} θ \\ =(a^{2} \sec^{2} θ - a^{2} \tan^{2} θ) - (b^{2} \sec^{2} θ - b^{2} \tan^{2} θ) \\ = a^{2} (\sec^{2} θ - \tan^{2} θ) - b^{2} (\sec^{2} θ - \tan^{2} θ) \\ = a^{2} - b^{2} [∵ \sec^{2} θ - \tan^{2} θ = 1] \\ Hence, x^{2} - y^{2} = a^{2} - b^{2} \end{array}$

Page No 628:

Answer:

$\begin{array}{l} We have (\frac{x}{a} \sin θ - \frac{y}{b} \cos θ) = 1 \\ Squaring both side, we have: \\ {(\frac{x}{a} \sin θ - \frac{y}{b} \cos θ)}^{2} = {(1)}^{2} \\ \Rightarrow (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ - 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 1 ... (i) \\ Again, (\frac{x}{a} \cos θ + \frac{y}{b} \sin θ) = 1 \\ Squaring both side, we get: \\ {(\frac{x}{a} \cos θ + \frac{y}{b} \sin θ)}^{2} = {(1)}^{2} \\ \Rightarrow (\frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ + 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 1 ... (ii) \\ Now, adding (i) and (ii), we get : \\ (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ - 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) + (\frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ + 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ + \frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ = 2 \\ \Rightarrow (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{x^{2}}{a^{2}} \cos^{2} θ) + (\frac{y^{2}}{b^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} (\sin^{2} θ + \cos^{2} θ) + \frac{y^{2}}{b^{2}} (\cos^{2} θ + \sin^{2} θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2 [∵ \sin^{2} θ + \cos^{2} θ = 1] \\ ∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2 \end{array}$

Page No 628:

Answer:

$\begin{array}{l} We have x = a \cos^{3} θ \\ => \frac{x}{a} = \cos^{3} θ ... (i) \\ Again, y = b \sin^{3} θ \\ => \frac{y}{b} = \sin^{3} θ ... (ii) \\ Now, LHS= {(\frac{x}{a})}^{\frac{2}{3}} + {(\frac{y}{b})}^{\frac{2}{3}} \\ = (\cos^{3} θ)^{\frac{2}{3}} + {(\sin^{3} θ)}^{\frac{2}{3}} [From (i) and (ii)] \\ {=cos}^{2} θ + \sin^{2} θ \\ = 1 \\ Hence, LHS = RHS \end{array}$

Page No 628:

Answer:

$x = b \sec^{3} θ \Rightarrow \sec^{3} θ = \frac{x}{b} \Rightarrow \sec θ = {(\frac{x}{b})}^{\frac{1}{3}} . . . . . (1)$
Also,
$y = a \tan^{3} θ \Rightarrow \tan^{3} θ = \frac{y}{a} \Rightarrow \tan θ = {(\frac{y}{a})}^{\frac{1}{3}} . . . . . (2)$
We know
$\sec^{2} θ - \tan^{2} θ = 1 \Rightarrow {[{(\frac{x}{b})}^{\frac{1}{3}}]}^{2} - {[{(\frac{y}{a})}^{\frac{1}{3}}]}^{2} = 1 [Using (1) and (2)] \Rightarrow {(\frac{x}{b})}^{\frac{2}{3}} - {(\frac{y}{a})}^{\frac{2}{3}} = 1$

Page No 628:

Answer:

$\begin{array}{l} We have (\tan θ + \sin θ) = m and (\tan θ - \sin θ) = n \\ Now, LHS= {(m^{2} - n^{2})}^{2} \\ = {[{(\tan θ + \sin θ)}^{2} - {(\tan θ - \sin θ)}^{2}]}^{2} \\ = {[(\tan^{2} θ + \sin^{2} θ + 2 \tan θ \sin θ) - (\tan^{2} θ + \sin^{2} θ - 2 \tan θ \sin θ)]}^{2} \\ = {[(\tan^{2} θ + \sin^{2} θ + 2 \tan θ \sin θ - \tan^{2} θ - \sin^{2} θ + 2 \tan θ \sin θ)]}^{2} \\ = {(4 \tan θ \sin θ)}^{2} \\ = 16 \tan^{2} θ \sin^{2} θ \\ = 16 \frac{\sin^{2} θ}{\cos^{2} θ} \sin^{2} θ \\ = 16 \frac{(1 - \cos^{2} θ) \sin^{2} θ}{\cos^{2} θ} \\ = 16 [\tan^{2} θ (1 - \cos^{2} θ)] \\ = 16 (\tan^{2} θ - \tan^{2} θ \cos^{2} θ) \\ = 16 (\tan^{2} θ - \frac{\sin^{2} θ}{\cos^{2} θ} \times \cos^{2} θ) \\ = 16 (\tan^{2} θ - \sin^{2} θ) \\ = 16 (\tan θ + \sin θ) (\tan θ - \sin θ) \\ = 16 m n [(\tan θ + \sin θ) (\tan θ - \sin θ) = m n] \\ ∴ (m^{2} - n^{2}) {(m^{2} - n^{2})}^{2} = 16 m n \end{array}$

Page No 628:

Answer:

$\begin{array}{l} We have (\cot θ + \tan θ) = m and (\sec θ - cos θ) = n \\ Now, m^{2} n = [{(\cot θ + \tan θ)}^{2} (\sec θ - cos θ)] \\ = [{(\frac{1}{\tan θ} + \tan θ)}^{2} (\frac{1}{\cos θ} - \cos θ)] \\ = \frac{{(1 + \tan^{2} θ)}^{2}}{\tan^{2} θ} \times \frac{(1 - \cos^{2} θ)}{\cos θ} \\ = \frac{\sec^{4} θ}{\tan^{2} θ} \times \frac{\sin^{2} θ}{\cos θ} \\ = \frac{\sec^{4} θ}{\frac{\sin^{2} θ}{\cos^{2} θ}} \times \frac{\sin^{2} θ}{\cos θ} \\ = \frac{\cos^{2} θ \times \sec^{4} θ}{\cos θ} \\ = \cos θ \sec^{4} θ \\ = \frac{1}{\sec θ} \times \sec^{4} θ = \sec^{3} θ \\ ∴ {(m^{2} n)}^{\frac{2}{3}} = {(\sec^{3} θ)}^{\frac{2}{3}} = \sec^{2} θ \end{array}$

$\begin{array}{l} Again, m n^{2} = [(\cot θ + \tan θ) {(\sec θ - cos θ)}^{2}] \\ = [(\frac{1}{\tan θ} + \tan θ) . {(\frac{1}{\cos θ} - \cos θ)}^{2}] \\ = \frac{(1 + \tan^{2} θ)}{\tan θ} \times \frac{{(1 - \cos^{2} θ)}^{2}}{\cos^{2} θ} \\ = \frac{\sec^{2} θ}{\tan θ} \times \frac{\sin^{4} θ}{\cos^{2} θ} \\ = \frac{\sec^{2} θ}{\frac{\sin θ}{\cos θ}} \times \frac{\sin^{4} θ}{\cos^{2} θ} \\ = \frac{\sec^{2} θ \times \sin^{3} θ}{\cos θ} \\ = \frac{1}{\cos^{2} θ} \times \frac{\sec^{3} θ}{\cos θ} = \tan^{3} θ \\ ∴ {(m n^{2})}^{\frac{2}{3}} = {(\tan^{3} θ)}^{\frac{2}{3}} = \tan^{2} θ \\ Now, {(m^{2} n)}^{\frac{2}{3}} - {(m n^{2})}^{\frac{2}{3}} \\ = \sec^{2} θ - \tan^{2} θ = 1 \\ = RHS \\ Hence proved . \end{array}$

Page No 628:

Answer:

$We have (\cos ec θ -sin θ) = a^{3} \begin{array}{l} => a^{3} = (\frac{1}{\sin θ} - \sin θ) \end{array} \begin{array}{l} => a^{3} = \frac{(1 - \sin^{2} θ)}{\sin θ} \end{array} = \frac{\cos^{2} θ}{\sin θ} ∴ a = \frac{\cos^{\frac{2}{3}} θ}{\sin^{\frac{1}{3}} θ} \begin{array}{l} Again, (\sec θ - \cos θ) = b^{3} \end{array} = > b^{3} = (\frac{1}{\cos θ} - \cos θ) = \frac{(1 - \cos^{2} θ)}{\cos θ} = \frac{\sin^{2} θ}{\cos θ} ∴ b = \frac{\sin^{\frac{2}{3}} θ}{\cos^{\frac{1}{3}} θ} \begin{array}{l} Now, LHS= a^{2} b^{2} (a^{2} + b^{2}) \end{array} = a^{4} b^{2} + a^{2} b^{4} = a^{3} (a b^{2}) + (a^{2} b) b^{3} \begin{array}{ll} = \frac{\cos^{2} θ}{\sin θ} \times [\frac{\cos^{\frac{2}{3}} θ}{\sin^{\frac{1}{3}} θ} \times \frac{\sin^{\frac{4}{3}} θ}{\cos^{\frac{2}{3}} θ}] + [\frac{\cos^{\frac{4}{3}} θ}{\sin^{\frac{2}{3}} θ} \times \frac{\sin^{\frac{2}{3}} θ}{\cos^{\frac{1}{3}} θ}] \times \frac{\sin^{2} θ}{\cos θ} \\ = \frac{\cos^{2} θ}{\sin θ} \times \sin θ + \cos θ \times \frac{\sin^{2} θ}{\cos θ} \end{array} {=cos}^{2} θ + \sin^{2} θ = 1 = RHS Hence proved .$

Page No 629:

Answer:

Given:

x = secA + sinA .....(1)

y = secA – sinA .....(2)

Adding (1) and (2), we get

$x + y = \sec A + \sin A + \sec A - \sin A \Rightarrow 2 \sec A = x + y \Rightarrow \sec A = \frac{x + y}{2} \Rightarrow \frac{1}{\sec A} = \frac{2}{x + y}$
$\Rightarrow \cos A = \frac{2}{x + y} . . . . . (3)$

Subtracting (2) from (1), we get

$x - y = \sec A + \sin A - \sec A + \sin A \Rightarrow 2 \sin A = x - y \Rightarrow \sin A = \frac{x - y}{2} . . . . . (4)$

We know

$\cos^{2} A + \sin^{2} A = 1 \Rightarrow {(\frac{2}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} = 1 [Using (3) and (4)]$

Page No 629:

Answer:

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

Page No 629:

Answer:

$\cos θ - \sin θ = \sqrt{2} \sin θ$
Squaring on both sides, we get
${(\cos θ - \sin θ)}^{2} = {(\sqrt{2} \sin θ)}^{2} \Rightarrow \cos^{2} θ + \sin^{2} θ - 2 \sin θ \cos θ = 2 \sin^{2} θ \Rightarrow \cos^{2} θ - \sin^{2} θ = 2 \sin θ \cos θ \Rightarrow (\cos θ - \sin θ) (\cos θ + \sin θ) = 2 \sin θ \cos θ [a^{2} - b^{2} = (a - b) (a + b)]$
$\Rightarrow \sqrt{2} \sin θ (\cos θ + \sin θ) = 2 \sin θ \cos θ (\cos θ - \sin θ = \sqrt{2} \sin θ) \Rightarrow \cos θ + \sin θ = \frac{2 \sin θ \cos θ}{\sqrt{2} \sin θ} \Rightarrow \cos θ + \sin θ = \sqrt{2} \cos θ$

Page No 629:

Answer:

$\sec θ - \tan θ = \sqrt{2} \tan θ$
Squaring on both sides, we get
${(\sec θ - \tan θ)}^{2} = {(\sqrt{2} \tan θ)}^{2} \Rightarrow \sec^{2} θ + \tan^{2} θ - 2 \sec θ \tan θ = 2 \tan^{2} θ \Rightarrow \sec^{2} θ - \tan^{2} θ = 2 \sec θ \tan θ \Rightarrow (\sec θ - \tan θ) (\sec θ + \tan θ) = 2 \sec θ \tan θ [a^{2} - b^{2} = (a - b) (a + b)]$
$\Rightarrow \sqrt{2} \tan θ (\sec θ + \tan θ) = 2 \sec θ \tan θ (\sec θ - \tan θ = \sqrt{2} \tan θ) \Rightarrow \sec θ + \tan θ = \frac{2 \sec θ \tan θ}{\sqrt{2} \tan θ} \Rightarrow \sec θ + \tan θ = \sqrt{2} \sec θ$

Page No 629:

Answer:

Given: $\sec θ + \tan θ = p . . . . . (1)$
We know
$\sec^{2} θ - \tan^{2} θ = 1 \Rightarrow (\sec θ - \tan θ) (\sec θ + \tan θ) = 1 \Rightarrow (\sec θ - \tan θ) p = 1 [From (1)] \Rightarrow \sec θ - \tan θ = \frac{1}{p} . . . . . (2)$
Adding (1) and (2), we get
$\sec θ + \tan θ + \sec θ - \tan θ = p + \frac{1}{p} \Rightarrow 2 \sec θ = \frac{p^{2} + 1}{p} \Rightarrow \sec θ = \frac{p^{2} + 1}{2 p} \Rightarrow \frac{1}{\sec θ} = \frac{2 p}{p^{2} + 1}$
$\Rightarrow \cos θ = \frac{2 p}{p^{2} + 1} . . . . . (3)$
Subtracting (2) from (1), we get
$\sec θ + \tan θ - \sec θ + \tan θ = p - \frac{1}{p} \Rightarrow 2 \tan θ = \frac{p^{2} - 1}{p} \Rightarrow \tan θ = \frac{p^{2} - 1}{2 p} . . . . . (4)$
Now,
$\sin θ = \tan θ \times \cos θ \Rightarrow \sin θ = \frac{p^{2} - 1}{2 p} \times \frac{2 p}{p^{2} + 1} [Using (3) and (4)] \Rightarrow \sin θ = \frac{p^{2} - 1}{p^{2} + 1}$

Page No 629:

Answer:

Given: $cosec A + \cot A = m . . . . . (1)$
We know
${cosec}^{2} A - \cot^{2} A = 1 \Rightarrow (cosec A - \cot A) (cosec A + \cot A) = 1 [a^{2} - b^{2} = (a - b) (a + b)] \Rightarrow (cosec A - \cot A) m = 1 [From (1)] \Rightarrow cosec A - \cot A = \frac{1}{m} . . . . . (2)$
Adding (1) and (2), we get
$cosec A + \cot A + cosec A - \cot A = m + \frac{1}{m} \Rightarrow 2 cosec A = \frac{m^{2} + 1}{m} \Rightarrow cosec A = \frac{m^{2} + 1}{2 m} \Rightarrow \frac{1}{cosec A} = \frac{2 m}{m^{2} + 1}$
$\Rightarrow \sin A = \frac{2 m}{m^{2} + 1} . . . . . (3)$
Subtracting (2) from (1), we get
$cosec A + \cot A - cosec A + \cot A = m - \frac{1}{m} \Rightarrow 2 \cot A = \frac{m^{2} - 1}{m} \Rightarrow \cot A = \frac{m^{2} - 1}{2 m} . . . . . (4)$
Now,
$\cos A = \sin A \times \cot A \Rightarrow \cos A = \frac{2 m}{m^{2} + 1} \times \frac{m^{2} - 1}{2 m} [From (3) and (4)] \Rightarrow \cos A = \frac{m^{2} - 1}{m^{2} + 1}$

Page No 629:

Answer:

Given: $\sec A - \tan A = x . . . . . (1)$
We know
$\sec^{2} A - \tan^{2} A = 1 \Rightarrow (\sec A + \tan A) (\sec A - \tan A) = 1 [a^{2} - b^{2} = (a - b) (a + b)] \Rightarrow (\sec A + \tan A) x = 1 [From (1)] \Rightarrow \sec A + \tan A = \frac{1}{x} . . . . . (2)$
Adding (1) and (2), we get
$\sec A - \tan A + \sec A + \tan A = x + \frac{1}{x} \Rightarrow 2 \sec A = \frac{x^{2} + 1}{x} \Rightarrow \sec A = \frac{x^{2} + 1}{2 x} . . . . . (3)$
Subtracting (1) from (2), we get
$\sec A + \tan A - \sec A + \tan A = \frac{1}{x} - x \Rightarrow 2 \tan A = \frac{1 - x^{2}}{x} \Rightarrow \tan A = \frac{1 - x^{2}}{2 x} . . . . . (4)$
Dividing (3) by (4), we get
$\frac{\frac{1 + x^{2}}{2 x}}{\frac{1 - x^{2}}{2 x}} = \frac{\sec A}{\tan A} \Rightarrow \frac{1 + x^{2}}{1 - x^{2}} = \frac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}} \Rightarrow \frac{1 + x^{2}}{1 - x^{2}} = \frac{1}{\sin A} \Rightarrow \frac{1 + x^{2}}{1 - x^{2}} = cosec A$

Page No 630:

Answer:

(b) 1
$\begin{array}{l} sin A + {sin}^{2} A = 1 \\ =>sin A = 1 - {sin}^{2} A \\ = > sin A = {cos}^{2} A (∵ 1 - {sin}^{2} A) \\ = > {sin}^{2} A = {cos}^{4} A (Squaring both sides) \\ =>1 - {cos}^{2} A = {cos}^{4} A \\ {=> cos}^{4} A + {cos}^{2} A = 1 \end{array}$

Page No 630:

Answer:

$\frac{1 + \tan^{2} A}{1 + \cot^{2} A} = \frac{\sec^{2} A}{{cosec}^{2} A} = \frac{\sin^{2} A}{\cos^{2} A} (\sec θ = \frac{1}{\cos θ} and cosec θ = \frac{1}{\sin θ}) = \tan^{2} A$

Hence, the correct answer is option (c).

Page No 631:

Answer:

(b) cos A

$\begin{array}{l} (sec A + \tan A) (1 - sin A) \\ = (\frac{1}{cos A} + \frac{sin A}{cos A}) (1 - sin A) \\ = (\frac{1 + sin A}{cos A}) (1 - sin A) \\ = (\frac{1 - {sin}^{2} A}{cos A}) \\ = (\frac{{cos}^{2} A}{cos A}) \\ = cos A \end{array}$

Page No 631:

Answer:

$(1 + \tan θ + \sec θ) (1 + \cot θ - cosec θ) = 1 + \cot θ - cosec θ + \tan θ + \tan θ \cot θ - \tan θ cosec θ + \sec θ + \sec θ \cot θ - \sec θ cosec θ = 1 + \frac{\cos θ}{\sin θ} - \frac{1}{\sin θ} + \frac{\sin θ}{\cos θ} + \tan θ \times \frac{1}{\tan θ} - \frac{\sin θ}{\cos θ} \times \frac{1}{\sin θ} + \frac{1}{\cos θ} + \frac{1}{\cos θ} \times \frac{\cos θ}{\sin θ} - \frac{1}{\cos θ \sin θ}$
$= 1 + \frac{\cos θ}{\sin θ} - \frac{1}{\sin θ} + \frac{\sin θ}{\cos θ} + 1 - \frac{1}{\cos θ} + \frac{1}{\cos θ} + \frac{1}{\sin θ} - \frac{1}{\cos θ \sin θ} = 2 + \frac{\cos θ}{\sin θ} + \frac{\sin θ}{\cos θ} - \frac{1}{\cos θ \sin θ} = 2 + \frac{\cos^{2} θ + \sin^{2} θ}{\cos θ \sin θ} - \frac{1}{\cos θ \sin θ}$
$= 2 + \frac{1}{\cos θ \sin θ} - \frac{1}{\cos θ \sin θ} (\cos^{2} θ + \sin^{2} θ = 1) = 2$

Hence, the correct answer is option (d).

Page No 631:

Answer:

$\sin θ - \cos θ = 0 \Rightarrow \sin θ = \cos θ \Rightarrow \frac{\sin θ}{\cos θ} = 1 \Rightarrow \tan θ = \tan 45 °$
$\Rightarrow θ = 45 ° ∴ \sin^{4} θ + \cos^{4} θ = \sin^{4} 45 ° + \cos^{4} 45 ° = {(\frac{1}{\sqrt{2}})}^{4} + {(\frac{1}{\sqrt{2}})}^{4} = \frac{1}{4} + \frac{1}{4}$
$= \frac{2}{4} = \frac{1}{2}$

Hence, the correct answer is option (b).

Page No 631:

Answer:

$\cos 9 α = \sin α \Rightarrow \cos 9 α = \cos (90 ° - α) \Rightarrow 9 α = 90 ° - α \Rightarrow 10 α = 90 °$
$\Rightarrow 5 α = \frac{90 °}{2} = 45 ° ∴ \tan 5 α = \tan 45 ° = 1$

Thus, the value of tan5α is 1.

Hence, the correct answer is option (c).

Page No 631:

Answer:

$\frac{\tan^{2} θ - \sec^{2} θ}{\cot^{2} θ - {cosec}^{2} θ} = \frac{- 1}{- 1} (1 + \tan^{2} θ = \sec^{2} θ and 1 + \cot^{2} θ = {cosec}^{2} θ) = 1$

Hence, the correct answer is option (d).

Page No 631:

Answer:

(b) (sec A − tan A)

$\sqrt{\frac{1 - sin A}{1 + sin A}} = \sqrt{\frac{(1 - sin A)}{(1 + sin A)} \times \frac{(1 - sin A)}{(1 - sin A)}} [Multiplying the denominator and numerator by (1 - sin A)] = \frac{(1 - sin A)}{\sqrt{1 - {sin}^{2} A}} = \frac{(1 + sin A)}{\sqrt{{cos}^{2} A}} = \frac{(1 - sin A)}{cos A} = \frac{1}{cos A} - \frac{sin A}{cos A} =sec A - \tan A$

Page No 631:

Answer:

$\sqrt{\frac{1 + \cos (A)}{1 - \cos (A)}} = \sqrt{\frac{1 + \cos (A)}{1 - \cos (A)} \times \frac{1 + \cos (A)}{1 + \cos (A)}} = \frac{1 + \cos (A)}{\sqrt{1 - \cos^{2} (A)}} = \frac{1 + \cos (A)}{\sqrt{\sin^{2} (A)}} = \frac{1 + \cos (A)}{\sin (A)} = \frac{1}{\sin (A)} + \frac{\cos (A)}{\sin (A)} = cosec (A) + \cot (A) .$
Hence, the correct answer is option B.

Page No 631:

Answer:

(b) $\frac{1 - cosθ}{1 + cosθ}$

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

Page No 631:

Answer:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)² = 5²
⇒ tan² θ + cot² θ + 2 tan θ cot θ = 25
⇒ tan² θ + cot² θ + 2 = 25 [∵ tan θ = $\frac{1}{\cot θ}$ ]
⇒ tan² θ + cot² θ = 25 − 2 = 23

Page No 631:

Answer:

(c) $\frac{b + a}{b - a}$
$Given: tan θ = \frac{a}{b} Now, \frac{(cos θ + sin θ)}{(cos θ - sin θ)} = \frac{(1 + tan θ)}{(1 - tan θ)} [Dividing the numerator and denominator by cos θ] = \frac{(1 + \frac{a}{b})}{(1 - \frac{a}{b})} = \frac{(\frac{b + a}{b})}{(\frac{b - a}{b})} = \frac{(b + a)}{(b - a)}$

Page No 631:

Answer:

(d) $\frac{3}{4}$
$= \frac{{cosec}^{2} θ - \sec^{2} θ}{{cosec}^{2} θ + \sec^{2} θ} = \frac{\sin^{2} θ (\frac{1}{\sin^{2} θ} - \frac{1}{\cos^{2} θ})}{\sin^{2} θ (\frac{1}{\sin^{2} θ} + \frac{1}{\cos^{2} θ})} [Multiplying the numerator and denominator by \sin^{2} θ] = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$
= $\frac{1 - \frac{1}{7}}{1 + \frac{1}{7}} = \frac{6}{8} = \frac{3}{4}$

Page No 632:

Answer:

$\tan θ = \frac{8}{15} \Rightarrow \frac{1}{\tan θ} = \frac{15}{8} \Rightarrow \cot θ = \frac{15}{8} . . . . . (1)$
Now,
${cosec}^{2} θ = 1 + \cot^{2} θ \Rightarrow {cosec}^{2} θ = 1 + {(\frac{15}{8})}^{2} [Using (1)] \Rightarrow {cosec}^{2} θ = 1 + \frac{225}{64} = \frac{289}{64} \Rightarrow cosec θ = \sqrt{\frac{289}{64}} = \frac{17}{8}$

Hence, the correct answer is option (a).

Page No 632:

Answer:

$5 \tan θ = 3 \Rightarrow 5 \times \frac{\sin θ}{\cos θ} = 3 \Rightarrow 5 \sin θ = 3 \cos θ . . . . . (1)$
$∴ \frac{5 \sin θ - \cos θ}{5 \sin θ + \cos θ} = \frac{3 \cos θ - \cos θ}{3 \cos θ + \cos θ} [Using (1)] = \frac{2 \cos θ}{4 \cos θ} = \frac{1}{2}$

Hence, the correct answer is option (c).

Page No 632:

Answer:

In ∆ABC,

∠A + ∠B + ∠C = 180º (Angle sum property of triangle)

⇒ ∠A + ∠B + 90º = 180º (∠C = 90º)

⇒ ∠A + ∠B = 180º − 90º = 90º

∴ cos(A + B) = cos90º = 0

Thus, the value of cos(A + B) is 0.

Hence, the correct answer is option (a).

Page No 632:

Answer:

(a) 1
$\begin{array}{l} cos A + {cos}^{2} A = 1 \\ =>cos A = 1 - {cos}^{2} A \\ =>cos A = {sin}^{2} A (∵ 1 - {cos}^{2} A = {sin}^{2}) \\ {=>cos}^{2} A = {sin}^{4} A (Squaring both sides) \\ =>1 - {sin}^{2} A = {sin}^{4} A \\ {=> sin}^{4} A + {sin}^{2} A = 1 \end{array}$

Page No 632:

Answer:

(d) 9

We have $\frac{(5 \sin θ + 3 \cos θ)}{(5 \sin θ - 3 \cos θ)}$ .

Dividing the numerator and denominator of the given expression by sin θ, we get:
$\frac{\frac{1}{\sin θ} (5 \sin θ + 3 \cos θ)}{\frac{1}{\sin θ} (5 \sin θ - 3 \cos θ)}$

= $\frac{5 + 3 c o t θ}{5 - 3 co t θ}$

= $\frac{5 + 4}{5 - 4}$ = 9 [∵ 3 cot θ = 4]

Page No 632:

Answer:

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = $\frac{\sec A}{2}$ and $\frac{1}{x} = \frac{\tan A}{2}$ .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2 $(x^{2} - \frac{1}{x^{2}})$ = 2 $({(\frac{\sec A}{2})}^{2} - {(\frac{\tan A}{2})}^{2})$
= 2 $((\frac{\sec^{2} A}{4}) - (\frac{\tan^{2} A}{4}))$
= $\frac{2}{4} (\sec^{2} A - \tan^{2} A)$
= $\frac{1}{2}$ [By using the identity: $(\sec^{2} θ - \tan^{2} θ = 1)$ ]

Page No 632:

Answer:

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = $\frac{\cos e c θ}{3}$ and $\frac{1}{x} = \frac{c o t θ}{3}$ .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3 $(x^{2} - \frac{1}{x^{2}})$ = 3 $({(\frac{\cos e c θ}{3})}^{2} - {(\frac{c o t θ}{3})}^{2})$
= 3 $((\frac{\cos e c^{2} θ}{9}) - (\frac{c o t^{2} θ}{9}))$
= $\frac{3}{9} (\cos e c^{2} θ - c o t^{2} θ)$
= $\frac{1}{3}$ [By using the identity: $(\cos e c^{2} θ - c o t^{2} θ = 1)$ ]

Page No 632:

Answer:

(d) 2

Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{B C}{A B}$
So, $\frac{B C}{A B}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$ k and AB = k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = ( $\sqrt{3}$ k)² + (k)²
⇒ AC²= 4k²
⇒ AC = 2k
∴ sec θ = $\frac{AC}{AB} = \frac{2 k}{k} = \frac{2}{1}$

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