Rs Aggarwal 2015 for Class 10 Math Chapter 7 - Trigonometric Identities

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Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 7 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 216:

Question 33:

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

Answer:

$\begin{array}{l} LHS= {\frac{1}{{sec}^{2} θ - {cos}^{2} θ} + \frac{1}{{cosec}^{2} θ - {sin}^{2} θ}} ({sin}^{2} θ {cos}^{2} θ) \\ = {\frac{{cos}^{2} θ}{1 - {cos}^{4} θ} + \frac{{sin}^{2} θ}{1 - {sin}^{4} θ}} ({sin}^{2} θ {cos}^{2} θ) \\ = {\frac{{cos}^{2} θ}{(1 - {cos}^{2} θ) ({1+cos}^{2} θ)} + \frac{\sin^{2} θ}{(1 - \sin^{2} θ) (1 + \sin^{2} θ)}} ({sin}^{2} θ {cos}^{2} θ) \\ = [\frac{\cot^{2} θ}{1 + \cos^{2} θ} + \frac{\tan^{2} θ}{1 + \sin^{2} θ}] \sin^{2} θ \cos^{2} θ \\ = \frac{\cos^{4} θ}{1 + \cos^{2} θ} + \frac{\sin^{4} θ}{1 + \sin^{2} θ} \\ = \frac{{(\cos^{2} θ)}^{2}}{1 + \cos^{2} θ} + \frac{{(\sin^{2} θ)}^{2}}{1 + \sin^{2} θ} \\ = \frac{(1 - \sin^{2} θ)}{1 + \cos^{2} θ} + \frac{{(1 - \cos^{2} θ)}^{2}}{1 + \sin^{2} θ} \\ = \frac{{(1 - \sin^{2} θ)}^{2} (1 + \sin^{2}) + {(1 - \cos^{2} θ)}^{2} (1 + {cos}^{2} θ)}{(1 + \sin^{2} θ) (1 + {cos}^{2} θ)} \\ = \frac{\cos^{4} θ (1 + \sin^{2} θ) + \sin^{4} θ (1 + {cos}^{2} θ)}{1 + \sin^{2} θ + \cos^{2} θ + \sin^{2} θ \cos^{2} θ} \end{array} = \frac{\cos^{4} θ + \cos^{4} θ \sin^{2} θ + \sin^{4} θ + \sin^{4} θ {cos}^{2} θ}{1 + 1 + \sin^{2} θ \cos^{2} θ} = \frac{\cos^{4} θ + \sin^{4} θ + \sin^{2} θ \cos^{2} θ (\sin^{2} θ + {cos}^{2} θ)}{2 + \sin^{2} θ \cos^{2} θ} = \frac{(\cos^{2} θ)^{2} + (\sin^{2} θ)^{2} + \sin^{2} θ \cos^{2} θ (1)}{2 + \sin^{2} θ \cos^{2} θ} = \frac{(\cos^{2} θ + \sin^{2} θ)^{2} - 2 \sin^{2} θ \cos^{2} θ + \sin^{2} θ \cos^{2} θ (1)}{2 + \sin^{2} θ \cos^{2} θ} \begin{array}{l} = \frac{1^{2} + \cos^{2} θ \sin^{2} θ - 2 \cos^{2} θ \sin^{2} θ}{2 + \sin^{2} θ \cos^{2} θ} \\ = \frac{1 - \cos^{2} θ \sin^{2} θ}{2 + \sin^{2} θ \cos^{2} θ} \end{array} = RHS$

Page No 216:

Question 34:

Prove each of the following identities:

$(i) \sqrt{\frac{1 + \sin θ}{1 - \sin θ}} = (\sec θ + \tan θ) (ii) \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = (cosec θ - \cot θ) (iii) \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ$

Answer:

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

$(ii) LHS = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = \sqrt{\frac{(1 - \cos θ)}{(1 + \cos θ)} \times \frac{(1 - \cos θ)}{(1 - \cos θ)}} = \sqrt{\frac{{(1 - \cos θ)}^{2}}{1 - \cos^{2} θ}} = \sqrt{\frac{{(1 - \cos θ)}^{2}}{\sin^{2} θ}} = \frac{1 - \cos θ}{\sin θ} = \frac{1}{\sin θ} - \frac{\cos θ}{\sin θ} = (cosec θ - \cot θ) = RHS$

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

Page No 216:

Question 35:

$\frac{(\sin A - \sin B)}{(\cos A + \cos B)} + \frac{(\cos A - \cos B)}{(\sin A + \sin B)} = 0$

Answer:

$\begin{array}{l} LHS= \frac{(\sin A - \sin B)}{(cos A + cos B)} + \frac{(cos A - cos B)}{(\sin A + \sin B)} \\ = \frac{(\sin A - \sin B) (\sin A + \sin B) + (cos A - cos B) (cos A - cos B)}{(cos A + cos B) (\sin A + \sin B)} \\ = \frac{{sin}^{2} A - {sin}^{2} B + {cos}^{2} A - {cos}^{2} B}{(cos A + cos B) (\sin A + \sin B)} \end{array} \begin{array}{l} = \frac{0}{(cos A + cos B) (\sin A + \sin B)} \\ =0 \\ =RHS \end{array}$

Page No 314:

Question 1:

(i) (1 − cos²θ) cosec²θ = 1
(ii) (1 + cot²θ) sin²θ = 1

Answer:

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

Page No 314:

Question 2:

(i) (sec²θ − 1) cot²θ = 1
(ii) (sec²θ − 1) (cosec²θ − 1) = 1
(iii) (1− cos²θ) sec²θ = tan²θ

Answer:

$\begin{array}{l} \begin{array}{l} (i) LHS = (\sec^{2} θ - 1) \cot^{2} θ \\ = \tan^{2} θ \times \cot^{2} θ (∵ \sec^{2} θ - \tan^{2} θ = 1) \\ = \frac{1}{\cot^{2} θ} \times \cot^{2} θ \\ =1 \\ =RH S \end{array} \\ (ii) LHS= (\sec^{2} θ - 1) ({cosec}^{2} θ - 1) \\ {=tan}^{2} θ \times \cot^{2} θ (∵ \sec^{2} θ - \tan^{2} θ = 1 a n d c o \sec^{2} θ - c o t^{2} θ = 1) \\ {=tan}^{2} θ \times \frac{1}{\tan^{2} θ} \\ =1 \\ =RH S \\ (iii) LHS = (1 - \cos^{2} θ) \sec^{2} θ \\ {=sin}^{2} θ \times \sec^{2} θ (∵ \sin^{2} θ + \cos^{2} θ = 1) \\ {=sin}^{2} θ \times \frac{1}{\cos^{2} θ} \\ = \frac{\sin^{2} θ}{\cos^{2} θ} \\ {=tan}^{2} θ \\ =RH S \end{array}$

Page No 314:

Question 3:

(i) $\sin^{2} θ + \frac{1}{(1 + \tan^{2} θ)} = 1$
(ii) $\frac{1}{(1 + \tan^{2} θ)} + \frac{1}{(1 + \cot^{2} θ)} = 1$

Answer:

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

Page No 314:

Question 4:

(i) (1 + cos θ) (1 − cos θ) (1 + cot²θ) = 1
(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1

Answer:

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

Page No 314:

Question 5:

(i) $1 + \frac{\cot^{2} θ}{(1 + cosecθ)} = cosecθ$
(ii) $1 + \frac{\tan^{2} θ}{(1 + secθ)} = secθ$

Answer:

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

Page No 314:

Question 6:

(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

Answer:

$\begin{array}{l} (i) LHS=sec θ (1 - sin θ) (sec θ + tan θ) \\ = (sec θ - sec θ sin θ) (sec θ + tan θ) \\ = (sec θ - \frac{1}{cos θ} \times sin θ) (sec θ + tan θ) \\ = (sec θ - tan θ) (sec θ + tan θ) \\ {=sec}^{2} θ - {tan}^{2} θ \\ =1 \\ =RHS \\ (ii) LHS=sin θ (1 + tan θ) + cos θ (1 + cot θ) \\ =sin θ + sin θ × \frac{sin θ}{cos θ} + cos θ +cos θ × \frac{cos θ}{sin θ} \\ = \frac{\cos θ {sin}^{2} θ + {sin}^{3} θ + {cos}^{2} θ sin θ {+cos}^{3} θ}{cos θ sin θ} \\ = \frac{({sin}^{3} θ + {cos}^{3} θ) + (cos θ {sin}^{2} θ + {cos}^{2} θ sin θ)}{cos θ sin θ} \\ = \frac{(sin θ + cos θ) ({sin}^{2} θ - sin θ cos θ {+cos}^{2} θ) + sin θ cos θ (sin θ +cos θ)}{cos θ sin θ} \\ = \frac{(sin θ +cos θ) ({sin}^{2} θ + {cos}^{2} θ - sin θ cos θ + sin θ cos θ)}{cos θ sin θ} \\ = \frac{(sin θ +cos θ) (1)}{cos θ sin θ} \\ = \frac{sin θ}{cos θ sin θ} + \frac{cos θ}{cos θ sin θ} \\ = \frac{1}{cos θ} + \frac{1}{sin θ} \\ =sec θ +cosec θ \\ =RHS \end{array}$

Page No 314:

Question 7:

$\frac{sinθ}{1 + cosθ} + \frac{(1 + cosθ)}{sinθ} = 2 cosecθ$

Answer:

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

Hence, LHS = RHS

Page No 314:

Question 8:

$\frac{tanθ}{(1 - cotθ)} + \frac{cotθ}{(1 - tanθ)} = (1 + secθ cosecθ)$

Answer:

$\begin{array}{l} LHS= \frac{tan θ}{(1 - \cot θ)} + \frac{cot θ}{(1 - tan θ)} \\ = \frac{tan θ}{(1 - \frac{cos θ}{sin θ})} + \frac{cot θ}{(1 - \frac{sin θ}{cos θ})} \\ = \frac{sin θ tan θ}{(sin θ - cos θ)} + \frac{cos θ cot θ}{(cos θ - sin θ)} \\ = \frac{sin θ × \frac{sin θ}{cos θ} - cos θ × \frac{cos θ}{sin θ}}{(sin θ - cos θ)} \\ = \frac{\frac{{sin}^{2} θ}{cos θ} - \frac{{cos}^{2} θ}{sin θ}}{(sin θ - cos θ)} \\ = \frac{{sin}^{3} θ - {cos}^{3} θ}{cos θ sin θ (sin θ - cos θ)} \\ = \frac{(sin θ - cos θ) ({sin}^{2} θ + sin θ cos θ {+cos}^{2} θ)}{cos θ sin θ (sin θ - cos θ)} \\ = \frac{1 + sin θ cos θ}{cos θ sin θ} \\ = \frac{1}{cos θ sin θ} + \frac{sin θ cos θ}{cos θ sin θ} \\ =sec θ cosec θ +1 \\ =1+sec θ cosec θ \\ =RHS \end{array}$

Page No 314:

Question 9:

$\frac{\cos^{2} θ}{(1 - tanθ)} + \frac{\sin^{3} θ}{(sinθ - cosθ)} = (1 + sinθ cosθ)$

Answer:

$\frac{\cos^{2} θ}{(1 - tanθ)} + \frac{\sin^{3} θ}{(sinθ - cosθ)} = (1 + sinθ cosθ)$

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

Hence, L.H.S. = R.H.S.

Page No 314:

Question 10:

$\frac{cosθ}{(1 - tanθ)} + \frac{\sin^{2} θ}{(cosθ - sinθ)} = (cosθ + sinθ)$

Answer:

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

Hence, LHS = RHS

Page No 314:

Question 11:

$\frac{\tan^{2} θ}{(1 + \tan^{2} θ)} + \frac{\cot^{2} θ}{(1 + \cot^{2} θ)} = 1$

Answer:

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

Hence, LHS = RHS

Page No 314:

Question 12:

$\frac{(1 + \tan^{2} θ) cotθ}{{cosec}^{2} θ} = \tan θ$

Answer:

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

Page No 314:

Question 13:

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

Answer:

$\begin{array}{l} LHS= (1 + {tan}^{2} θ) (1 + {cot}^{2} θ) \\ {=sec}^{2} θ . {cosec}^{2} θ (\begin{array}{l} ∵ {sec}^{2} θ - {tan}^{2} θ = 1 and \\ {cosec}^{2} θ - {cot}^{2} θ = 1 \end{array}) \\ = \frac{1}{{cos}^{2} θ . {sin}^{2} θ} \\ = \frac{1}{(1 - {sin}^{2} θ) {sin}^{2} θ} \\ = \frac{1}{{sin}^{2} θ - {sin}^{4} θ} \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 315:

Question 14:

$\frac{tanθ}{(1 + \tan^{2} θ)^{2}} + \frac{cotθ}{(1 + \cot^{2} θ)^{2}} = sinθ cosθ$

Answer:

$\begin{array}{l} LHS= \frac{tan θ}{{(1 + {tan}^{2} θ)}^{2}} + \frac{cot θ}{{(1 + {cot}^{2} θ)}^{2}} \\ = \frac{tan θ}{{({sec}^{2} θ)}^{2}} + \frac{cot θ}{{({cosec}^{2} θ)}^{2}} \\ = \frac{tan θ}{{sec}^{4} θ} + \frac{cot θ}{{cosec}^{4} θ} \\ = \frac{sin θ}{cos θ} \times {cos}^{4} θ + \frac{cos θ}{sin θ} \times {sin}^{4} θ \\ =sin θ {cos}^{3} θ + cos θ {sin}^{3} θ \\ =sin θ cos θ ({cos}^{2} θ + {sin}^{2} θ) \\ =sin θ cos θ \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 315:

Question 15:

(i) $\sin^{6} θ + \cos^{6} θ = 1 - 3 \sin^{2} {θcos}^{2} θ$
(ii) $\sin^{2} θ + \cos^{4} θ = \cos^{2} θ + \sin^{4} θ$
(iii) ${cosec}^{4} θ - {cosec}^{2} θ = \cot^{4} θ + \cot^{2} θ$

Answer:

$\begin{array}{l} (i) {LHS=sin}^{6} θ + {cos}^{6} θ \\ = {({sin}^{2} θ)}^{3} + {({cos}^{2} θ)}^{3} \\ = ({sin}^{2} θ + {cos}^{2} θ) ({sin}^{4} θ - {sin}^{2} θ {cos}^{2} θ + {cos}^{4} θ) \\ =1×{{({sin}^{2} θ)}^{2} + 2 {sin}^{2} θ {cos}^{2} θ + {({cos}^{2} θ)}^{2} - 3 {sin}^{2} θ {cos}^{2} θ} \\ = {({sin}^{2} θ + {cos}^{2} θ)}^{2} - 3 {sin}^{2} θ {cos}^{2} θ \\ = {(1)}^{2} - 3 {sin}^{2} θ {cos}^{2} θ \\ =1 - 3 {sin}^{2} θ {cos}^{2} θ \\ =RHS \end{array} Hence, LHS = RHS \begin{array}{l} (ii) {LHS=sin}^{2} θ + {cos}^{4} θ \\ {=sin}^{2} θ + {({cos}^{2} θ)}^{2} \\ {=sin}^{2} θ + {(1 - {sin}^{2} θ)}^{2} \\ {=sin}^{2} θ + 1 - 2 {sin}^{2} θ + {sin}^{4} θ \\ =1 - {sin}^{2} θ + {sin}^{4} θ \\ {=cos}^{2} θ + {sin}^{4} θ \\ =RHS \end{array} Hence, LHS = RHS \begin{array}{l} (iii) {LHS=cosec}^{4} θ - {cosec}^{2} θ \\ {=cosec}^{2} θ ({cosec}^{2} θ - 1) \\ {=cosec}^{2} θ \times {cot}^{2} θ (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ = (1 + {cot}^{2} θ) {×cot}^{2} θ \\ {=cot}^{2} θ + {cot}^{4} θ \\ =RHS \end{array} Hence, LHS = RHS$

Page No 315:

Question 16:

(i) $\frac{1 - sinθ}{1 + sinθ} = (secθ - tanθ)^{2}$
(ii) $\frac{1 + cosθ}{1 - cosθ} = (cosecθ + cotθ)^{2}$

Answer:

$\begin{array}{l} (i) LHS= \frac{1 - sin θ}{1+sin θ} \\ = \frac{(1 - sin θ) (1 - sin θ)}{(1+sin θ) (1 - sin θ)} [\begin{array}{l} Dividing the numerator and \\ denominator by (1 - sin θ) \end{array}] \\ = \frac{{(1 - sin θ)}^{2}}{(1 - {sin}^{2} θ)} \\ = \frac{1 + {sin}^{2} θ - 2 sin θ}{{cos}^{2} θ} \\ = \frac{1}{{cos}^{2} θ} + \frac{{sin}^{2} θ}{{cos}^{2} θ} - \frac{2 sin θ}{{cos}^{2} θ} \\ {=sec}^{2} θ + {tan}^{2} θ - 2 \times \frac{1}{cos θ} \times \frac{sin θ}{cos θ} \\ {=sec}^{2} θ + {tan}^{2} θ - 2 sec θ tan θ \\ = {(sec θ - tan θ)}^{2} \\ =RHS \end{array} Hence, LHS = RHS \begin{array}{l} (ii) LHS= \frac{1 + cos θ}{1 - cos θ} \\ = \frac{(1 + cos θ) (1 + cos θ)}{(1 - cos θ) (1 + cos θ)} [\begin{array}{l} Dividing the numerator and \\ denominator by (1 + cos θ) \end{array}] \\ = \frac{{(1 + cos θ)}^{2}}{(1 - {cos}^{2} θ)} \\ = \frac{1 + {cos}^{2} θ + 2 cos θ}{{sin}^{2} θ} \\ = \frac{1}{{sin}^{2} θ} + \frac{{cos}^{2} θ}{{sin}^{2} θ} + \frac{2 cos θ}{{sin}^{2} θ} \\ {=cosec}^{2} θ + {cot}^{2} θ + 2 \frac{1}{sin θ} \times \frac{cos θ}{sin θ} \\ {=cosec}^{2} θ + {cot}^{2} θ + 2 {cosec}^{2} θ {cot}^{2} θ \\ = {(cosec θ + cot θ)}^{2} \\ =RHS \end{array} Hence, LHS = RHS$

Page No 315:

Question 17:

$\frac{1 + \tan^{2} θ}{1 + \cot^{2} θ} = {(\frac{1 + tanθ}{1 - cotθ})}^{2} = \tan^{2} θ$

Answer:

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

∴ LHS = RHS
Hence proved.

Page No 315:

Question 18:

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

Answer:

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

Page No 315:

Question 19:

(i) $\frac{tanθ}{(secθ - 1)} + \frac{tanθ}{(secθ + 1)} = 2 cosecθ$
(ii) $\frac{cotθ}{(cosecθ + 1)} + \frac{(cosecθ + 1)}{cotθ} = 2 secθ$

Answer:

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

Page No 315:

Question 20:

(i) $\frac{secθ - 1}{secθ + 1} = \frac{\sin^{2} θ}{(1 + cosθ)^{2}}$
(ii) $\frac{secθ - tanθ}{secθ + tanθ} = \frac{\cos^{2} θ}{(1 + sinθ)^{2}}$

Answer:

$\begin{array}{l} (i) LHS= \frac{sec θ - 1}{sec θ +1} \\ = \frac{\frac{1}{cos θ} - 1}{\frac{1}{cos θ} + 1} \\ = \frac{\frac{1 - cos θ}{cos θ}}{\frac{1 + cos θ}{cos θ}} \\ = \frac{1 - cos θ}{1 + cos θ} \\ = \frac{(1 - cos θ) (1 + cos θ)}{(1 + cos θ) (1 + cos θ)} {\begin{cases} Dividing the numerator and \\ denominator by (1 + cos θ) \end{cases}} \\ = \frac{1 - {cos}^{2} θ}{{(1 + cos θ)}^{2}} \\ = \frac{{sin}^{2} θ}{{(1 + cos θ)}^{2}} \\ =RHS \\ (ii) LHS= \frac{sec θ - tan θ}{sec θ +tan θ} \\ = \frac{\frac{1}{cos θ} - \frac{sin θ}{cos θ}}{\frac{1}{cos θ} + \frac{sin θ}{cos θ}} \\ = \frac{\frac{1 - sin θ}{cos θ}}{\frac{1+sin θ}{cos θ}} \\ = \frac{1 - sin θ}{1+sin θ} \\ = \frac{(1 - sin θ) (1+sin θ)}{(1+sin θ) (1+sin θ)} {\begin{array}{l} Dividing the numerator and \\ denominator by (1+sin θ) \end{array}} \\ = \frac{(1 - {sin}^{2} θ)}{{(1+sin θ)}^{2}} \\ = \frac{{cos}^{2} θ}{{(1+sin θ)}^{2}} \\ =RHS \end{array}$

Page No 315:

Question 21:

$\frac{sinθ}{(cotθ + cosecθ)} - \frac{sinθ}{(cotθ - cosecθ)} = 2$

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 315:

Question 22:

(i) $\frac{sinθ - cosθ}{sinθ + cosθ} + \frac{sinθ + cosθ}{sinθ - cosθ} = \frac{2}{(2 \sin^{2} θ - 1)}$
(ii) $\frac{sinθ + cosθ}{sinθ - cosθ} + \frac{sinθ - cosθ}{sinθ + cosθ} = \frac{2}{(1 - 2 \cos^{2} θ)}$

Answer:

$\begin{array}{l} (i) LHS= \frac{sin θ - cos θ}{sin θ + cos θ} + \frac{sin θ + cos θ}{sin θ - cos θ} \\ = \frac{{(sin θ - cos θ)}^{2} + {(sin θ + cos θ)}^{2}}{(sin θ + cos θ) (sin θ - cos θ)} \\ = \frac{{sin}^{2} θ + {cos}^{2} θ - 2 sin θ cos θ + {sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{1 + 1}{{sin}^{2} θ - (1 - {sin}^{2} θ)} (∵ {sin}^{2} θ + {cos}^{2} θ = 1) \\ = \frac{2}{{sin}^{2} θ - 1 + {sin}^{2} θ} \\ = \frac{2}{2 {sin}^{2} θ - 1} \\ =RHS \\ (ii) LHS= \frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} \\ = \frac{{(sin θ + cos θ)}^{2} + {(sin θ - cos θ)}^{2}}{(sin θ - cos θ) (sin θ + cos θ)} \\ = \frac{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ + {sin}^{2} θ + {cos}^{2} θ - 2 sin θ cos θ}{({sin}^{2} θ - {cos}^{2} θ)} \\ = \frac{1 + 1}{(1 - {cos}^{2} θ) - {cos}^{2} θ} (∵ {sin}^{2} θ + {cos}^{2} θ = 1) \\ = \frac{2}{1 - 2 {cos}^{2} θ} \\ =RHS \end{array}$

Page No 315:

Question 23:

$\frac{\cos^{3} θ + \sin^{3} θ}{cosθ + sinθ} + \frac{\cos^{3} θ - \sin^{3} θ}{cosθ - sinθ} = 2$

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 315:

Question 24:

$(1 - sinθ + cosθ)^{2} = 2 (1 + cosθ) (1 - sinθ)$

Answer:

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

Hence, LHS = RHS

Page No 315:

Question 25:

$\frac{1 + cosθ - \sin^{2} θ}{sinθ (1 + cosθ)} = cotθ$

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 316:

Question 26:

(i) $\frac{cosecθ + cotθ}{cosecθ - cotθ} = (cosecθ + cotθ)^{2} = 1 + 2 \cot^{2} + 2 cosecθ cotθ$
(ii) $\frac{secθ + tanθ}{secθ - tanθ} = (secθ + tanθ)^{2} = 1 + 2 \tan^{2} θ + 2 secθ tanθ$

Answer:

$\begin{array}{l} (i) Here, \frac{cosec θ + cot θ}{cosec θ - cot θ} \\ = \frac{(cosec θ + cot θ) (cosec θ + cot θ)}{(cosec θ - cot θ) (cosec θ + cot θ)} \\ = \frac{{(cosec θ + cot θ)}^{2}}{({cosec}^{2} θ - {cot}^{2} θ)} \\ = \frac{{(cosec θ + cot θ)}^{2}}{1} \\ = {(cosec θ + cot θ)}^{2} \\ Again, {(cosec θ + cot θ)}^{2} \\ {=cosec}^{2} θ + {cot}^{2} θ + 2 cosec θ cot θ \\ {=1+cot}^{2} θ + {cot}^{2} θ + 2 cosec θ cot θ (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ {=1+2cot}^{2} θ + 2 cosec θ cot θ \\ (ii) Here, \frac{sec θ +tan θ}{sec θ - tan θ} \\ = \frac{(sec θ +tan θ) (sec θ +tan θ)}{(sec θ - tan θ) (sec θ +tan θ)} \\ = \frac{{(sec θ +tan θ)}^{2}}{{sec}^{2} θ - {tan}^{2} θ} \\ = \frac{{(sec θ +tan θ)}^{2}}{1} \\ = {(sec θ +tan θ)}^{2} \\ Again, {(sec θ +tan θ)}^{2} \\ {=sec}^{2} θ + {tan}^{2} θ + 2 sec θ tan θ \\ {=1+tan}^{2} θ + {tan}^{2} θ + 2 sec θ tan θ \\ {=1+2tan}^{2} θ + 2 sec θ tan θ \end{array}$

Page No 316:

Question 27:

(i) $\frac{1 + cosθ + sinθ}{1 + cosθ - sinθ} = \frac{1 + sinθ}{cosθ}$
(ii) $\frac{sinθ + 1 - cosθ}{cosθ - 1 + sinθ} = \frac{1 + sinθ}{cosθ}$

Answer:

$\begin{array}{l} (i) LHS= \frac{1 + cos θ +sin θ}{1 + cos θ - sin θ} \\ = \frac{{(1 + cos θ) + sin θ} {(1 + cos θ) + sin θ}}{{(1 + cos θ) - sin θ} {(1 + cos θ) + sin θ}} {Multiplying the numerator and denominator by (1 + cos θ + sinθ)} \\ = \frac{{(1 + cos θ) + sin θ}^{2}}{{{(1 + cos θ)}^{2} - {sin}^{2} θ}} \\ = \frac{1 + {cos}^{2} θ + 2 cos θ {+sin}^{2} θ + 2 sin θ (1 + cos θ)}{1 + {cos}^{2} θ + 2 cos θ - {sin}^{2} θ} \\ = \frac{2 + 2 cos θ + 2 sin θ (1 + cos θ)}{1 + {cos}^{2} θ + 2 cos θ - (1 - {cos}^{2} θ)} \\ = \frac{2 (1 + cos θ) + 2 sin θ (1 + cos θ)}{2 {cos}^{2} θ + 2 cos θ} \\ = \frac{2 (1 + cos θ) (1 + sin θ)}{2 cos θ (1 + cos θ)} \\ = \frac{1 + sin θ}{cos θ} \\ =RHS \\ (ii) LHS= \frac{sin θ +1 - cos θ}{cos θ - 1 + sin θ} \\ = \frac{(sin θ +1 - cos θ) (sin θ +cos θ +1)}{(cos θ - 1 + sin θ) (sin θ +cos θ +1)} {Multiplying and dividing by 1 + cos θ + sinθ } \\ = \frac{{(sin θ + 1)}^{2} - {cos}^{2} θ}{{(sin θ +cos θ)}^{2} - 1^{2}} \\ = \frac{{sin}^{2} θ + 1 + 2 sin θ - {cos}^{2} θ}{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ - 1} \\ = \frac{{sin}^{2} θ + {sin}^{2} θ + {cos}^{2} θ + 2 sin θ - {cos}^{2} θ}{2 sin θ cos θ} \\ = \frac{2 {sin}^{2} θ + 2 sin θ}{2 sin θ cos θ} \\ = \frac{2 sin θ (1 + sin θ)}{2sin θ cos θ} \\ = \frac{1+sin θ}{cos θ} \\ =RHS \end{array}$

Page No 316:

Question 28:

$\frac{sinθ}{(secθ + tanθ - 1)} + \frac{cosθ}{(cosecθ + cotθ - 1)} = 1$

Answer:

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

Page No 316:

Question 29:

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

Answer:

$\begin{array}{l} We have \frac{sin θ +cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ +cos θ} \\ = \frac{{(sin θ +cos θ)}^{2} + {(sin θ - cos θ)}^{2}}{(sin θ - cos θ) (sin θ +cos θ)} \\ = \frac{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ + {sin}^{2} θ {+cos}^{2} θ - 2 sin θ cos θ}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{1 + 1}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{2}{{sin}^{2} θ - {cos}^{2} θ} \\ Again, \frac{2}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{2}{{sin}^{2} θ - (1 - {sin}^{2} θ)} \\ = \frac{2}{2 {sin}^{2} θ - 1} \end{array}$

Page No 316:

Question 30:

$\frac{cosθ cosecθ - sinθ secθ}{cosθ + sinθ} = cosecθ - secθ$

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 316:

Question 31:

$(1 + tanθ + cotθ) (sinθ - cosθ) = (\frac{secθ}{{cosec}^{2} θ} - \frac{cosecθ}{\sec^{2} θ})$

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 316:

Question 32:

$\frac{\cot^{2} θ (secθ - 1)}{(1 + sinθ)} = \sec^{2} θ . (\frac{1 - sinθ}{1 + secθ})$

Answer:

$\begin{array}{l} LHS= \frac{{cot}^{2} θ (sec θ - 1)}{(1 + sin θ)} \\ = \frac{\frac{{cos}^{2} θ}{{sin}^{2} θ} (\frac{1}{cos θ} - 1)}{(1 + sin θ)} \\ = \frac{\frac{cos θ}{{sin}^{2} θ} - \frac{{cos}^{2} θ}{{sin}^{2} θ}}{(1 + sin θ)} \\ = \frac{cos θ (1 - cos θ)}{{sin}^{2} θ (1 + sin θ)} \\ = \frac{cos θ (1 - cos θ)}{(1 - {cos}^{2} θ) (1 + sin θ)} \\ = \frac{cos θ (1 - cos θ)}{(1+cos θ) (1 - cos θ) (1 + sin θ)} \\ = \frac{cos θ}{(1 + \frac{1}{sec θ}) (1 + sin θ)} \\ = \frac{cos θ sec θ}{(sec θ +1) (1 + sin θ)} \\ = \frac{1}{(sec θ +1) (1 + sin θ)} \\ = \frac{(1 - sin θ)}{(sec θ +1) (1 + sin θ) (1 - sin θ)} [Multiplying the numerator and denominator by (1-sinθ)] \\ = \frac{(1 - sin θ)}{(sec θ +1) (1 - \sin^{2} θ)} \\ = \frac{(1 - sin θ)}{(sec θ +1) {cos}^{2} θ} \\ = \frac{{sec}^{2} θ (1 - sin θ)}{(sec θ +1)} \\ =RHS \end{array}$
Hence, LHS = RHS

Page No 316:

Question 36:

$\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B$

Answer:

$\begin{array}{l} LHS= \frac{\tan A + \tan B}{\cot A + \cot B} \\ = \frac{\tan A + \tan B}{\frac{1}{\tan A} + \frac{1}{\tan B}} \\ = \frac{\tan A + \tan B}{\frac{\tan A + \tan B}{\tan A \tan B}} \\ = \frac{\tan A \tan B (\tan A + \tan B)}{(t an A + \tan B)} \\ = \tan A \tan B \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 316:

Question 37:

Show that none of the following is an identity:
(i) cos²θ + cos θ = 1
(ii) sin2θ + sin θ = 2
(iii) tan²θ + sin θ = cos²θ

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 317:

Question 38:

Show that each of the following is an identity:
(i) $\frac{\tan^{2} θ}{(1 + \tan^{2} θ)} = \sin^{2} θ$
(ii) $\frac{cotθ + cosθ}{cotθ - cosθ} = \frac{1 + sinθ}{1 - sinθ}$

Answer:

$\begin{array}{l} (i) LHS= \frac{{tan}^{2} θ}{(1 + {tan}^{2} θ)} \\ = \frac{{tan}^{2} θ}{{sec}^{2} θ} \\ = \frac{\frac{{sin}^{2} θ}{\cos^{2} θ}}{\frac{1}{\cos^{2} θ}} \\ {=sin}^{2} θ \\ =RHS \end{array} i.e., LHS=RHS ∴ T his is an identity . \begin{array}{l} (ii) LHS= \frac{cot θ +cos θ}{\cot θ - \cos θ} \\ = \frac{\frac{\cos θ}{\sin θ} + \cos θ}{\frac{\cos θ}{\sin θ} - \cos θ} \\ = \frac{\cos θ + \cos θ \sin θ}{\cos θ - \cos θ \sin θ} \\ = \frac{\cos θ (1 + \sin θ)}{\cos θ (1 - \sin θ)} \\ = \frac{1 + \sin θ}{1 - \sin θ} \\ =RHS \\ i . e ., LHS = RHS \end{array} ∴ T his is an identity .$

Page No 321:

Question 1:

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that (m² + n²) = (a² + b²).

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 321:

Question 2:

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x² − y²) = (a² − b²).

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 322:

Question 3:

$If (\frac{x}{a} sinθ - \frac{y}{b} cosθ) = 1 and (\frac{x}{a} cosθ + \frac{y}{b} sinθ) = 1,$ prove that $(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = 2 .$

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 322:

Question 4:

If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

Answer:

$\begin{array}{l} We have (\sec θ + \tan θ) = m ... (i) \\ Again, (\sec θ - \tan θ) = n ... (ii) \\ Now, multiplying (i) and (ii), we get : \end{array} \begin{array}{l} (\sec θ + \tan θ) \times (\sec θ - \tan θ) = m n \\ = > \sec^{2} θ - \tan^{2} θ = m n \\ = > 1 = m n [∵ \sec^{2} θ - \tan^{2} θ = 1] \\ ∴ m n = 1 \end{array}$

Page No 322:

Question 5:

If (cosec θ + cot θ) = m and (cosec θ − cot θ) = n, show that mn = 1.

Answer:

$\begin{array}{l} We have (cosec θ + \cot θ) = m ... (i) \\ Again, (cosec θ - \cot θ) = n ... (ii) \\ Now, multiplying (i) and (ii), we get: \end{array} \begin{array}{l} (cosec θ + \cot θ) \times (cosec θ - \cot θ) = m n \\ = > \cos {ec}^{2} θ - \cot^{2} θ = m n \\ = > 1 = m n [∵ \cos {ec}^{2} θ - \cot^{2} θ = 1] \\ ∴ m n = 1 \end{array}$

Page No 322:

Question 6:

If x = a cos³θ and y = b sin³θ, prove that ${(\frac{x}{a})}^{2 / 3} + {(\frac{y}{b})}^{2 / 3} = 1 .$

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 322:

Question 7:

If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m²− n²)² = 16mn.

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 322:

Question 8:

If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m²n)^2/3 − (mn²)^2/3 = 1.

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 322:

Question 9:

If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m²n)^2/3 − (mn²)^2/3 = 1.

Answer:

Page No 322:

Question 10:

If (cos θ + sin θ) = 1, prove that (cos θ − sin θ) = ± 1.

Answer:

$We have (\cos θ + \sin θ) = 1 W e know that {(\cos θ + \sin θ)}^{2} + {(\cos θ - \sin θ)}^{2} = 2 (\sin^{2} θ + \cos^{2} θ) = > {(1)}^{2} + {(\cos θ - \sin θ)}^{2} = 2 \times 1 = > 1 + {(\cos θ - \sin θ)}^{2} = 2 = > {(\cos θ - \sin θ)}^{2} = 1 = > \cos θ - \sin θ = \pm 1 ∴ \cos θ - \sin θ = \pm 1$

Page No 322:

Question 11:

If tan A = n tan B and sin A = m sin B, prove that cos²A = $\frac{(m^{2} - 1)}{n^{2} - 1} .$

Answer:

$\begin{array}{l} We have \tan A = n \tan B \end{array} => \cot B = \frac{n}{\tan A} ... (i) Again, \sin A = m \sin B = > \cos ec B = \frac{m}{\sin A} ... (ii) S quaring (i) and (ii) and subtracting (ii) from (i), we get \frac{m^{2}}{\sin^{2} A} - \frac{n^{2}}{\tan^{2} A} = \cos e c^{2} B - \cot^{2} B = > \frac{m^{2}}{\sin^{2} A} - \frac{n^{2} \cos}{\sin^{2} A} = 1 = > m^{2} - n^{2} \cos^{2} A = \sin^{2} A = > m^{2} - n^{2} \cos^{2} A = 1 - \cos^{2} A = > n^{2} \cos^{2} A - \cos^{2} A = m^{2} - 1 = > \cos^{2} A (n^{2} - 1) = (m^{2} - 1) = > \cos^{2} A = \frac{(m^{2} - 1)}{(n^{2} - 1)} ∴ \cos^{2} A = \frac{(m^{2} - 1)}{(n^{2} - 1)}$

Page No 322:

Question 12:

If (cosec θ − sin θ) = a³and (sec θ − cos θ) = b³, prove that $a^{2} b^{2} (a^{2} + b^{2}) = 1 .$

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 322:

Question 13:

If a cos³θ + 3a sin²θ cos θ = m and a sin³θ + 3a sin θ cos²θ = n, prove that
(m + n)^2/3 + (m − n)^2/3 = 2a²^/3.

Answer:

$We have a \cos^{3} θ + 3 a \sin^{2} θ \cos θ = m and a \sin^{3} θ + 3 a \sin θ \cos^{2} θ = n Now, (m + n) = a \cos^{3} θ + 3 a \sin^{2} θ \cos θ + a \sin^{3} θ + 3 a \sin θ \cos^{2} θ = a [(\cos^{3} θ + \sin^{3} θ) + 3 \sin θ \cos θ (\sin θ + \cos θ)] \Rightarrow (\frac{m + n}{a}) = [(\cos^{3} θ + \sin^{3} θ) + 3 \sin θ \cos θ (\sin θ + \cos θ)] \Rightarrow m + n = {a (\cos θ + \sin θ)}^{3} ∴ {(m + n)}^{\frac{2}{3}} = {(\cos θ + \sin θ)}^{2} a^{\frac{2}{3}} ... (i) Similarly, {(m - n)}^{\frac{2}{3}} = {(\cos θ - \sin θ)}^{2} a^{\frac{2}{3}} ... (ii) Adding equation (i) and (ii), we get : {(m + n)}^{\frac{2}{3}} + {(m - n)}^{\frac{2}{3}} = a^{\frac{2}{3}} [\cos^{2} θ + \sin^{2} θ + 2 \sin θ \cos θ + \cos^{2} θ + \sin^{2} θ - 2 \sin θ \cos θ] = a^{\frac{2}{3}} [2] = 2 a^{\frac{2}{3}}$

Page No 322:

Question 14:

If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ± 3.

Answer:

$Given, (2 \sin θ + 3 \cos θ) = 2 ... (i) \begin{array}{l} We have {(2 \sin θ + 3 \cos θ)}^{2} + {(3 \sin θ - 2 \cos θ)}^{2} \end{array} {=4sin}^{2} θ + 9 \cos^{2} θ + 12 \sin θ \cos θ + 9 \sin^{2} θ + 4 \cos^{2} θ - 12 \sin θ \cos θ =4 (\sin^{2} θ + \cos^{2} θ) + 9 (\sin^{2} θ + \cos^{2} θ) =4+9 =13 i . e ., {(2 \sin θ + 3 \cos θ)}^{2} + {(3 \sin θ - 2 \cos θ)}^{2} = 13 = > 2^{2} + {(3 \sin θ - 2 \cos θ)}^{2} = 13 = > {(3 \sin θ - 2 \cos θ)}^{2} = 13 - 4 = > {(3 \sin θ - 2 \cos θ)}^{2} = 9 = > (3 \sin θ - 2 \cos θ) = \pm 3$

Page No 322:

Question 15:

If sin θ + cos θ = $\sqrt{2}$ sin (90° − θ), show that cot θ = $(\sqrt{2} + 1)$ .

Answer:

$Given: cos θ +sin θ = \sqrt{2} \sin (90^{0} - θ) =>cos θ +sin θ = \sqrt{2} \cos θ {∵ sin(90°-θ) = cosθ} => \frac{cos θ +sin θ}{\sin θ} = \frac{\sqrt{2} \cos θ}{\sin θ} (Dividing both sides by sin θ) =>cot θ + 1 = \sqrt{2} \cot θ => (\sqrt{2} - 1) \cot θ = 1 =>cot θ = \frac{1}{(\sqrt{2} - 1)} =>cot θ = \frac{1}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)} \begin{array}{l} =>cot θ = \frac{(\sqrt{2} + 1)}{2 - 1} \end{array} =>cot θ = (\sqrt{2} + 1) Hence proved .$

Page No 322:

Question 16:

If cos θ + sin θ = $\sqrt{2}$ sin θ, show that sin θ − cos θ = $\sqrt{2}$ cos θ.

Answer:

$Given: cos θ +sin θ = \sqrt{2} \sin θ We have {(\sin θ + \cos θ)}^{2} + {(\sin θ - \cos θ)}^{2} = 2 (\sin^{2} θ + \cos^{2} θ) => {(\sqrt{2} \sin θ)}^{2} + {(\sin θ - \cos θ)}^{2} = 2 {=>2sin}^{2} θ + {(\sin θ - \cos θ)}^{2} = 2 => {(\sin θ - \cos θ)}^{2} = 2 - 2 \sin^{2} θ => {(\sin θ - \cos θ)}^{2} = 2 (1 - \sin^{2} θ) => {(\sin θ - \cos θ)}^{2} = 2 \cos^{2} θ => (\sin θ - \cos θ) = \sqrt{2} \cos θ Hence proved .$

Page No 322:

Question 17:

If tan θ = $\frac{a}{b}$ , show that $\frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ .

Answer:

$Given: tan θ = \frac{a}{b} LHS= \frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ} = \frac{a \tan θ - b}{a \tan θ + b} (\begin{array}{l} Dividing both numerator and \\ denominator by cos θ \end{array}) = \frac{a \times \frac{a}{b} - b}{a \times \frac{a}{b} + b} = \frac{\frac{a^{2}}{b} - b}{\frac{a^{2}}{b} + b} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

Hence, LHS = RHS

Page No 323:

Question 1:

If sin A + sin² A = 1, then (cos² A + cos⁴ A) = ?
(a) $\frac{1}{2}$
(b) 1
(c) 2
(d) 3

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 324:

Question 2:

If cos A + cos² A = 1, then (sin² A + sin⁴ A) = ?
(a) 1
(b) 2
(c) 4
(d) 3

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 324:

Question 3:

(sec A + tan A) (1 − sin A) = ?
(a) sin A
(b) cos A
(c) sec A
(d) cosec A

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 324:

Question 4:

(1 + tan A + sec A)(1 + cot A − cosec A) = ?
(a) 1
(b) 0
(c) 2
(d) 4

Answer:

(c) 2

$(1 + \tan A + s ec A) (1 + c ot A - c o \sec A) = (1 + \frac{sin A}{cos A} + \frac{1}{cos A}) (1 + \frac{cos A}{sin A} - \frac{1}{sin A}) = (\frac{cos A + sin A + 1}{cos A}) (\frac{sin A + c os A - 1}{sin A}) = \frac{{(cos A + sin A) + 1} {(sin A + cos A) - 1}}{cos A sin A} = \frac{{(cos A + s in A)}^{2} - 1}{cos A sin A} = \frac{({cos}^{2} A + {sin}^{2} A + 2 cos A sin A) - 1}{cos A sin A} = \frac{1 + 2 cos A sin A - 1}{cos A sin A} (∵ {cos}^{2} A + {sin}^{2} A = 1) = \frac{2 cos A sin A}{cos A sin A} = 2$

Page No 324:

Question 5:

$\frac{(1 + \tan^{2} A)}{(1 + \cot^{2} A)} = ?$
(a) tan2 A
(b) cot2 A
(c) sec2 A
(d) cosec2 A

Answer:

(a) tan² A
$\frac{1 + \tan^{2} A}{1 + \cot^{2} A} = \frac{1 + \frac{{sin}^{2} A}{{cos}^{2} A}}{1 + \frac{{cos}^{2} A}{{sin}^{2} A}} = \frac{\frac{{cos}^{2} A + s {in}^{2} A}{{cos}^{2} A}}{\frac{{sin}^{2} A + c {os}^{2} A}{{sin}^{2} A}} = \frac{\frac{1}{{cos}^{2} A}}{\frac{1}{{sin}^{2} A}} (∵ c {os}^{2} A + s {in}^{2} A = 1) = \frac{{sin}^{2} A}{{cos}^{2} A} = {tan}^{2} A$

Page No 324:

Question 6:

(sin A + cos A)² + (sin A − cos A)² = ?
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(c) 2
${(s in A + cos A)}^{2} + {(sin A - cos A)}^{2} = ({sin}^{2} A + 2 sin A cos A + {cos}^{2} A) + ({sin}^{2} A + {cos}^{2} A - 2 sin A cos A) = 2 {sin}^{2} A + 2 {cos}^{2} A = 2 ({sin}^{2} A + {cos}^{2} A) = 2 \times 1 (∵ {sin}^{2} A + {cos}^{2} A = 1) = 2$

Page No 324:

Question 7:

(sec⁴ A − sec² A) = ?
(a) tan⁴ A − tan² A
(b) tan² A + tan⁴ A
(c) tan² A − tan⁴ A
(d) None of these

Answer:

(b) tan² A + tan⁴ A

$({sec}^{4} A - {sec}^{2} A) = {sec}^{2} A ({sec}^{2} A - 1) = {sec}^{2} A ({tan}^{2} A) = (1 + {tan}^{2} A) ({tan}^{2} A) = {tan}^{2} A + {tan}^{4} A$

Page No 324:

Question 8:

(cos⁴θ − sin⁴θ)=?
(a) 1 − 2 sin²θ
(b) 1 − 2 cos²θ
(c) 2 − sin²θ
(d) 2 − cos²θ

Answer:

a) 1 − 2 sin²θ

$({cos}^{4} θ - {sin}^{4} θ) = {({cos}^{2} θ)}^{2} - {({sin}^{2} θ)}^{2} = ({cos}^{2} θ + {sin}^{2} θ) ({cos}^{2} θ - {sin}^{2} θ) = 1 (1 - {sin}^{2} θ - {sin}^{2} θ) = 1 - 2 {sin}^{2} θ$

Page No 324:

Question 9:

$\frac{sinθ}{(1 + cosθ)} = ?$
(a) $\frac{cosθ}{(1 - sinθ)}$
(b) $\frac{(1 - cosθ)}{sinθ}$
(c) $\frac{(1 - sinθ)}{cosθ}$
(d) None of these

Answer:

(b) $\frac{(1 - cosθ)}{sinθ}$

$\frac{sin θ}{1 + c os θ} = \frac{sin θ (1 - c os θ)}{(1 + c os θ) (1 - c os θ)} [\begin{array}{l} Multiplying the numerator and \\ denominator by (1 - cos θ) \end{array}] = \frac{sin θ (1 - cos θ)}{1 - {cos}^{2} θ} = \frac{sin θ (1 - cos θ)}{{sin}^{2} θ} = \frac{(1 - cos θ)}{sin θ}$

Page No 324:

Question 10:

$\frac{sinθ}{(1 + cosθ)} + \frac{sinθ}{(1 - cosθ)} = ?$
(a) 2 sin θ
(b) 2 cos θ
(c) 2 sec θ
(d) 2 cosec θ

Answer:

(d) 2 cosec θ

$(\frac{sin θ}{1 + c os θ}) + (\frac{sin θ}{1 - c os θ}) = \frac{sin θ (1 - c os θ) + s in θ (1 + c os θ)}{(1 + c os θ) (1 - c os θ)} = \frac{sin θ - s in θ c os θ +sin θ +sin θ c os θ}{1 - c {os}^{2} θ} = \frac{2 s in θ}{{sin}^{2} θ} [∵ 1 - c {os}^{2} θ = s {in}^{2} θ] = \frac{2}{sin θ} = 2 c osec θ$

Page No 324:

Question 11:

$\frac{sinθ}{(1 - cotθ)} + \frac{cosθ}{(1 - tanθ)} = ?$
(a) (cos θ + sin θ)
(b) (cos θ − sin θ)
(c) 0
(d) 2 tan θ

Answer:

(a) (cos θ + sin θ)

$\frac{sin θ}{(1 - c ot θ)} + \frac{cos θ}{(1 - tan θ)} = \frac{sin θ}{1 - \frac{cos θ}{sin θ}} + \frac{cos θ}{1 - \frac{sin θ}{cos θ}} = (\frac{sin θ}{\frac{sin θ - c os θ}{sin θ}}) + (\frac{cos θ}{\frac{cos θ - s in θ}{cos θ}}) = (\frac{{sin}^{2} θ}{sin θ - c os θ}) + (\frac{{cos}^{2} θ}{cos θ - s in θ}) = (\frac{{cos}^{2} θ}{cos θ - s in θ}) - (\frac{{sin}^{2} θ}{cos θ - s in θ}) = (\frac{{cos}^{2} θ - s {in}^{2} θ}{cos θ - s in θ}) = {\frac{(c os θ +sin θ) (c os θ - s in θ)}{(c os θ - s in θ)}} =cos θ + s in θ$

Page No 324:

Question 12:

$\frac{(1 - \tan^{2} θ)}{(1 + \tan^{2} θ)} = ?$
(a) (sin²θ − cos²θ)
(b) (cos²θ − sin²θ)
(c) (cot²θ − tan²θ)
(d) (tan²θ − cot²θ)

Answer:

(b) (cos²θ − sin²θ)

$\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ} = \frac{(1 - \frac{{Sin}^{2} θ}{{Cos}^{2} θ})}{(1 + \frac{{Sin}^{2} θ}{{Cos}^{2} ϑ})} = \frac{(\frac{{Cos}^{2} θ - {Sin}^{2} θ}{{Cos}^{2} θ})}{(\frac{{Cos}^{2} θ + {Sin}^{2} θ}{{Cos}^{2} θ})} = \frac{{cos}^{2} θ - s {in}^{2} θ}{{cos}^{2} θ + s {in}^{2} θ} {=cos}^{2} θ - s {in}^{2} θ [∵ c {os}^{2} θ + s {in}^{2} θ = 1]$

Page No 325:

Question 13:

$\sqrt{\frac{1 + \sin A}{1 - \sin A}} = ?$
(a) (sec A + tan A)
(b) (sec A − tan A)
(c) sec A tan A
(d) None of these

Answer:

(a) (sec A + tan A)

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

Page No 325:

Question 14:

$\sqrt{\frac{1 - \sin A}{1 + \sin A}} = ?$
(a) (sec A + tan A)
(b) (sec A − tan A)
(c) sec A tan A
(d) None to these

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 325:

Question 15:

$\sqrt{\frac{1 - \cos A}{1 + \cos A}} = ?$
(a) (cosec A − cot A)
(b) (cosec A + cot A)
(c) cosec A cot A
(d) None of these

Answer:

(a) (cosec A − cot A)

$\sqrt{\frac{1 - cos A}{1 + \cos A}} = \sqrt{\frac{(1 - cos A)}{(1 + \cos A)} \times \frac{(1 - cos A)}{(1 - \cos A)}} [\begin{array}{l} Multiplying the numerator and \\ denominator by (1 - \cos A) \end{array}] = \sqrt{\frac{(1 - cos A) (1 - cos A)}{1 - {cos}^{2} A}} = \frac{1 - co s A}{\sqrt{1 - {cos}^{2} A}} = \frac{1 - co s A}{\sqrt{{sin}^{2} A}} = \frac{1 - co s A}{\sin A} = \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \cos ec A - cot A$

Page No 325:

Question 16:

$\sqrt{\frac{1 + \cos A}{1 - \cos A}} = ?$
(a) (cosec A − cot A)
(b) (cosec A + cot A)
(c) cosec A cot A
(d) None of these

Answer:

(b) (cosec A + cot A)

$\sqrt{\frac{1 + cos A}{1 - \cos A}} = \sqrt{\frac{(1 + cos A)}{(1 - \cos A)} \times \frac{(1 + cos A)}{(1 + \cos A)}} [\begin{array}{l} Multiplying the numerator and \\ denominator by (1 + \cos A) \end{array}] = \sqrt{\frac{(1 + cos A) (1 + cos A)}{1 - {cos}^{2} A}} = \frac{1 + co s A}{\sqrt{1 - {cos}^{2} A}} = \frac{1 + co s A}{\sqrt{{sin}^{2} A}} = \frac{1 + co s A}{\sin A} = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \cos ec A + cot A$

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Question 17:

$\sqrt{\frac{s e c A - \tan A}{s e c A + \tan A}} = ?$
(a) (sec A − tan A)
(b) (sec A + tan A)
(c) sec A tan A
(d) None of these

Answer:

(a) (sec A − tan A)

$\sqrt{\frac{\sec A - \tan A}{\sec A + \tan A}} By rationalising, we get : \sqrt{\frac{(\sec A - \tan A)}{(\sec A + \tan A)} \times \frac{(\sec A - \tan A)}{(\sec A - \tan A)}} = \sqrt{\frac{(\sec A - \tan A) (\sec A - \tan A)}{(\sec^{2} A - \tan^{2} A)}} = \frac{(\sec A - \tan A)}{1} [∵ {sec}^{2} A - {tan}^{2} A = 1] = \sec A - \tan A$

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Question 18:

(sin⁴θ − cos⁴θ + 1) cosec²θ = ?
(a) 4
(b) 3
(c) 2
(d) 1

Answer:

(c) 2

$({sin}^{4} θ - {cos}^{4} θ + 1) {cosec}^{2} θ = {({sin}^{2} θ - {cos}^{2} θ) ({sin}^{2} θ + {cos}^{2} θ) + 1} \frac{1}{{sin}^{2} θ} = [{({sin}^{2} θ - {cos}^{2} θ) \times 1} + 1] \frac{1}{{sin}^{2} θ} = \frac{{sin}^{2} θ + (1 - {cos}^{2} θ)}{{sin}^{2} θ} = \frac{{sin}^{2} θ + \sin^{2} θ}{{sin}^{2} θ} [∵ (1 - {cos}^{2} θ = {sin}^{2} θ)] = \frac{2 {sin}^{2} θ}{sin θ} = 2$

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Question 19:

If x = a cos θ and y = b sin θ, then (b²x² + a²y²) = ?
(a) a² + b²
(b) a²b²
(c) ab
(d) a⁴b⁴

Answer:

(b) a²b²

$x = a cos θ and y = b sin θ =>cos θ = \frac{x}{a} and sin θ = \frac{y}{b} {We know that cos}^{2} θ + {sin}^{2} θ = 1 => {(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} = 1 => \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 => \frac{x^{2} b^{2} + y^{2} a^{2}}{a^{2} b^{2}} = 1 => b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$

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Question 20:

If x = a sec θ cos ϕ, y = b θ sin ϕ and z = c tan θ, then $(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = ?$
(a) $(1 + \frac{z^{2}}{c^{2}})$
(b) $(1 - \frac{z^{2}}{c^{2}})$
(c) $(\frac{z^{2}}{c^{2}} - 1)$
(d) $\frac{z^{2}}{c^{2}}$

Answer:

(a) $(1 + \frac{z^{2}}{c^{2}})$

$x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = ctan θ ∴ sec θ cos ϕ = \frac{x}{a}, sec θ sin ϕ = \frac{y}{b} and tan θ = \frac{z}{c} {Now, sec}^{2} θ {cos}^{2} ϕ + {sec}^{2} θ {sin}^{2} ϕ = {(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} [Squaring both sides] {=>sec}^{2} θ ({cos}^{2} ϕ + {sin}^{2} ϕ) = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} {=>sec}^{2} θ = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =>1 + {tan}^{2} θ = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =>1+ \frac{z^{2}}{c^{2}} = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} ∴ (\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = (1+ \frac{z^{2}}{c^{2}})$

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Question 21:

$\sqrt{\frac{secθ - 1}{secθ + 1}} + \sqrt{\frac{secθ + 1}{secθ - 1}} = ?$
(a) 2 sin θ
(b) 2 cos θ
(c) 2 cosec θ
(d) 2 sec θ

Answer:

(c) 2 cosec θ

$\sqrt{\frac{sec θ - 1}{sec θ +1}} + \sqrt{\frac{sec θ +1}{sec θ - 1}} By rationalising, we get : \sqrt{\frac{(sec θ - 1) (sec θ - 1)}{(sec θ +1) (sec θ - 1)}} + \sqrt{\frac{(sec θ +1) (sec θ + 1)}{(sec θ - 1) (sec θ + 1)}} = \sqrt{\frac{(sec θ - 1) (sec θ - 1)}{({sec}^{2} θ - 1)}} + \sqrt{\frac{(sec θ +1) (sec θ + 1)}{({sec}^{2} θ - 1)}} = \frac{\sqrt{(sec θ - 1)^{2}}}{\sqrt{{tan}^{2} θ}} + \frac{\sqrt{(sec θ +1)^{2}}}{\sqrt{{tan}^{2} θ}} = \frac{(sec θ - 1)}{tan θ} + \frac{(sec θ +1)}{tan θ} = \frac{sec θ - 1 + sec θ +1}{tan θ} = \frac{2 sec θ}{tan θ} = \frac{2 \frac{1}{cos θ}}{\frac{sin θ}{cos θ}} =2cosec θ$

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Question 22:

If (sin θ + cos θ) = p and (sec θ + cosec θ) = q, then q(p² − 1) = ?
(a) 2
(b) 2p
(c) $\frac{1}{p^{2}}$
(d) $\frac{1}{p}$

Answer:

(b) 2p

$Given: sinθ + cosθ = p Squaring both sides, we get : {(sin θ +cos θ)}^{2} = p^{2} ... (i) and (sec θ +cosec θ) = q ... (i i) F rom (i), ({sin}^{2} θ + {cos}^{2} θ) + 2 sin θ cos θ = p^{2} =>1+ 2 sin θ cos θ = p^{2} => 2 sin θ cos θ = p^{2} - 1 =>sin θ cos θ = \frac{p^{2} - 1}{2} ... (iii) According to the question, we have: (sec θ +cosec θ) = q => (\frac{1}{cos θ} + \frac{1}{sin θ}) = q => (\frac{sin θ +cos θ}{\cos θ sin θ}) = q Hence, (\frac{p}{\frac{p^{2} - 1}{2}}) = q => \frac{2 p}{p^{2} - 1} = q => q (p^{2} - 1) =2 p$

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Question 23:

If (cos θ + sin θ) = $\sqrt{2}$ cos θ, then (cos θ − sin θ) = ?
(a) $\sqrt{2}$ sin θ
(b) $\sqrt{2}$ sec θ
(c) $\sqrt{2}$ cosec θ
(d) None of these

Answer:

(a) $\sqrt{2}$ sin θ

$Given: (cos θ +sin θ) = \sqrt{2} cos θ We have: {(cos θ +sin θ)}^{2} + {(cos θ - sin θ)}^{2} = 2 ({cos}^{2} θ {+sin}^{2} θ) => {(\sqrt{2} cos θ)}^{2} + {(cos θ - sin θ)}^{2} = 2 \times 1 => {(cos θ - sin θ)}^{2} = 2 - 2 {cos}^{2} θ => {(cos θ - sin θ)}^{2} = 2 (1 - {cos}^{2} θ) => {(cos θ - sin θ)}^{2} = 2 {sin}^{2} θ ∴ (cos θ - sin θ) = \sqrt{2} sin θ$

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Question 24:

$\frac{(sinθ - 2 \sin^{3} θ)}{(2 \cos^{3} θ - cosθ)} = ?$
(a) tan θ
(b) cot θ
(c) sec θ
(d) cosec θ

Answer:

(a) tan θ
$(\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ}) = \frac{sin θ (1 - 2 {sin}^{2} θ)}{cos θ (2 {cos}^{2} θ - 1)} = \frac{sin θ}{cos θ} × \frac{cos2 θ}{cos2 θ} [\begin{array}{l} ∵ (1 - 2 {sin}^{2} θ) = cos2 θ \\ (2 {cos}^{2} θ - 1) = cos2 θ \end{array}] =tan θ$

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Question 25:

(cosec θ − cot θ)² = ?
(a) $\frac{1 + cosθ}{1 - cosθ}$
(b) $\frac{1 - cosθ}{1 + cosθ}$
(c) $\frac{1 + sinθ}{1 - sinθ}$
(d) None of these

Answer:

(b) $\frac{1 - cosθ}{1 + cosθ}$

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

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Question 26:

If tan θ = $\frac{a}{b},$ then $\frac{(cosθ + sinθ)}{(cosθ - sinθ)} = ?$
(a) $\frac{a + b}{a - b}$
(b) $\frac{a - b}{a + b}$
(c) $\frac{b + a}{b - a}$
(d) $\frac{b - a}{b + a}$

Answer:

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

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Question 27:

If (sin θ + cos θ) = $\sqrt{3}$ , then (tan θ + cot θ) = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\frac{1}{2}$
(c) 1
(d) $\frac{3}{2}$

Answer:

(c) 1

$(sin θ + cos θ) = \sqrt{3} => {(sin θ + cos θ)}^{2} = 3 [Squaring both sides] {=>sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ = 3 =>1+ 2 sin θ cos θ = 3 => 2 sin θ cos θ = 2 =>sin θ cos θ = 1 ∴ sin θ cos θ = {sin}^{2} θ + {cos}^{2} θ => \frac{{sin}^{2} θ + {cos}^{2} θ}{sin θ cos θ} = 1 => \frac{sin θ}{cos θ} + \frac{cos θ}{sin θ} = 1 =>tan θ + cot θ = 1$

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Question 28:

If (sin θ + cos θ) = a and (sin³θ + cos³θ) = b, then (3a − 2b) = ?
(a) a³
(b) b³
(c) 0
(d) 1

Answer:

(a) a³

$Given: (sin θ + cos θ) = a \Rightarrow {(sin θ + cos θ)}^{2} = a^{2} [Squaring both sides] \Rightarrow ({sin}^{2} θ + {cos}^{2} θ) + 2 sin θ cos θ = a^{2} \Rightarrow 1+ 2 sin θ cos θ = a^{2} \Rightarrow sin θ cos θ = \frac{a^{2} - 1}{2} Again, {(sin θ + cos θ)}^{3} = a^{3} [C ubing both sides] \Rightarrow {sin}^{3} θ + {cos}^{3} θ + 3 sin θ cos θ (sin θ + \cos θ) = a^{3} ∴ b + 3 (\frac{a^{2} - 1}{2}) a = a^{3} \Rightarrow b + \frac{3 a (a^{2} - 1)}{2} = a^{3} \Rightarrow 2 b + 3 a (a^{2} - 1) = 2 a^{3} \Rightarrow 2 b + 3 a^{3} - 3 a = 2 a^{3} \Rightarrow - 3 a + 2 b = - a^{3} \Rightarrow 3 a - 2 b = a^{3}$

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