KC Sinha: Exercise 6.2 - Mathematics Solution Class 11 Chapter 6 त्रिकोणमितीय फलन

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(i) साबित करे कि (Prove that) sin65°+cos65°=√2cos20°
Sol :
L.H.S
=sin65°+cos65°
=sin65°+sin(90°-65°)
=sin65°+sin25°

∵$\left[\begin{array}{c} \sin C+\sin D =2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\end{array}\right]$

साबित करे कि (Prove that) $\frac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}$
Sol :
L.H.S
=$\frac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}$

$=\frac{\sin (90^{\circ}-10^{\circ})-\sin 10^{\circ}}{\sin (90^{\circ}-10^{\circ})+\sin 10^{\circ}}$

∵$\left[\begin{array}{c} \sin C-\sin D =2 cos \frac{C+D}{2} sin \frac{C-D}{2}\end{array}\right]$

∵$\left[\begin{array}{c} \sin C+\sin D =2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\end{array}\right]$

 $=\frac{2 \cos \frac{80^{\circ}+10^{\circ}}{2} \sin \frac{80^{\circ}-10^{\circ}}{2}}{2 \sin \frac{80^{\circ}+10^{\circ}}{2} \cos \frac{80^{\circ}-10^{\circ}}{2}}$

साबित करे कि (Prove that) cos80°+cos40°-cos20°=0
Sol :
L.H.S
=cos80°+cos40°-cos20°

$=2 Cos\frac{ 80^{\circ}+40^{\circ}}{2} \cos \frac{80^{\circ}-40^{\circ}}{2}-\cos 20^{\circ}$

साबित करे कि (Prove that) sin10°+sin20°+sin40°+sin50°=sin70°+sin80°
Sol :
L.H.S
=sin10°+sin20°+sin40°+sin50°

 $=2 \sin \frac{50^{\circ}+40^{\circ}}{2} \cdot \cos \frac{50^{\circ}-10^{\circ}}{2}+2 \sin \frac{40^{\circ}+20^{\circ}}{2} cos \frac{40^{\circ}-20^{\circ}}{2}$

$=2 \times \dfrac{1}{2} \times \cos 20^{\circ}+2 \times \dfrac{1}{2} \cos 10^{\circ}$

साबित करे कि (Prove that) $\cos \frac{\pi}{5}+\cos \frac{2 \pi}{5}+\cos \frac{6 \pi}{5}+\cos \frac{7 \pi}{5}=0$

$=\cos \frac{\pi}{5}+\cos \frac{2 \pi}{5}+\cos \left(\pi+\frac{\pi}{5}\right)+\cos \left(\pi+\frac{2 \pi}{5}\right)$

$=\cos \frac{\pi}{5}+\cos \frac{2 \pi}{5}-\cos \frac{\pi}{5}-\cos \frac{2 \pi}{5}$

∵$\left[\begin{array}{c} \sin C-\sin D =2 cos \frac{C+D}{2} sin \frac{C-D}{2}\end{array}\right]$

∵$\left[\begin{array}{c} \sin C+\sin D =2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\end{array}\right]$

$=\frac{2 cos \frac{7 x+5x}{2} \cos \frac{7 x-5 x}{2}}{2 \cos \frac{7 x+5}{2} \sin \frac{7 x-5}{2}}$

$=\frac{\cos \frac{2 x}{2}}{\tan \frac{2 x}{2}}$

$=\frac{\cos x}{\sin x}=\cot x$

(iv) cot 4x(six5x+sin3x)=cotx(sin5x-sin3x)
Sol :
L.H.S
=cot 4x(six5x+sin3x)

$=\cot 4 x \cdot 2 \sin \frac{5 x+3 x}{2} \cos \frac{5 x-3 x}{2}$

$=\frac{\cos x}{\sin 4 x} \times 2 \sin 4 x \times \cos x$

=2cos4x.cosx

R.H.S
=cotx(sin5x-sin3x)

$=\cot x \cdot 2 \cos \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}$

$=\dfrac{\cos x}{\sin x} \times 2 \cos 4 x \sin x$

=2cos4x.cosx

(v) $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
Sol :
L.H.S
$\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$

$=\frac{-(\sin 3 x-\sin x)}{-\left(\cos ^{2} x-\sin ^{2} x\right)}$

∵ cos²A-sin²B=cos(A+B).cos(A-B)

∵ cos²x-sin²x=cos2x

$=\frac{2 \cos \frac{3 x+x}{2} \sin \frac{3 x-x}{2}}{\cos (x+x) \cos (x-x)}$

$=\frac{2 \cos 2 x \sin x}{\cos 2 x \cos 0}$

$=\dfrac{2 \sin x}{1}$

=2sinx

Question 7

साबित करे कि (Prove that)

(i) $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$

Sol :

L.H.S

$\frac{\sin x-\sin y}{\cos x+\cos y}$

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

$=\tan \dfrac{x - y}{2}$

(ii) $\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \left(\frac{x+y}{2}\right) \cdot \cot \left(\frac{x-y}{2}\right)$

Sol :

Question 8

साबित करे कि (Prove that) 

(i) sin2x+2sin4x+sin6x=4cos²x six4x

Sol :

=sin2x+2sin4x+sin6x

=2sin4x+(sin6x+sin2x)

=$2 \sin 4 x+2 \sin \frac{6 x+2 x}{2} \cos \frac{6 x-2 x}{2}$

=2sin4x+2sin4x.cos2x

=2sin4x(1+cos2x)

[∵ 1+cos2x=2cos^2x]

=4cos²x.sin4x

(ii) sin x+sin 3x+sin 5x+sin 7x=4 cos x cos 2x sin 4 x

Sol :
L.H.S
= sin x+sin 3x+sin 5x+sin 7x

=(sin7x+sinx)+(sin5x+sin3x)

=$=2 \sin \frac{7 x+x}{2} \cos \frac{7 x-x}{2}+2 \sin \frac{5 x+3 x}{2} \cos \frac{5 x-3 x}{2}$

=2sin4x.cox3x+2sin4x.cosx

=2sin4x(cos3x+cosx)

=$=2sin 4 x \times 2 \cos \frac{3 x+1}{2} \cos \frac{3 x-x}{2}$

=4sin4x.cos2x.cosx

=4cosx.cos2x.sin4x

Question 9

साबित करे कि (Prove that) 

(i) $\frac{\cos 4 \theta+\cos 3 \theta+\cos 2 \theta}{\sin 4 \theta+\sin 3 \theta+\sin 2 \theta}=\cot 3 \theta$

Sol :

L.H.S

=$\dfrac{\cos 4 \theta+\cos 3 \theta+\cos 2 \theta}{\sin 4 \theta+\sin 3 \theta+\sin 2 \theta}$

$=\dfrac{(\cos 4 \theta+\cos 2 \theta)+\cos 3 \theta}{(\sin 4 \theta+\sin 2 \theta)+\sin 3 \theta}$

$=\dfrac{2 \cos \frac{4 \theta+2 \theta}{2} \cos \frac{4 \theta-2 \theta}{2}+\cos 3 \theta}{2 \sin \frac{4 \theta+2 \theta}{2} \cos \frac{4 \theta-2 \theta}{2}+\sin 3 \theta}$

$=\dfrac{2 \cos 3 \theta \cos \theta+\cos 3 \theta}{2 \sin 3 \theta \cos \theta+\sin 3 \theta}$

$=\frac{\cos 3 \theta\left(2 \cos \theta+1\right)}{\sin 3 \theta(2 \cos \theta+1)}$

=cot3θ

(ii) $\dfrac{\sin 5 \theta-2 \sin 3 \theta+\sin \theta}{\cos 5 \theta-\cos \theta}=\tan \theta$

Sol :

L.H.S

=$\dfrac{\sin 5\theta-2 \sin 3 \theta+\sin \theta}{\cos 5 \theta-\cos \theta}$

$=\frac{(\sin 5 \theta+\sin \theta)-2 \sin 3 \theta}{\cos 5 \theta-\cos \theta}$

∵$\left[\begin{array}{c} \cos C-\cos D =-2sin \frac{C+D}{2} sin \frac{C-D}{2}\end{array}\right]$

$=\frac{2 \tan \frac{5 \theta+\theta}{2} \cos \frac{5 \theta-\theta}{2}-2 \sin 3 \theta}{-2 \sin \frac{5 \theta+\theta}{2} \sin \frac{5 \theta-\theta}{2}}$

$=\frac{2 \sin 3 \theta \cos 2 \theta-2 \sin 3 \theta}{-2 \sin 3 \theta \sin 2 \theta}$

$=\frac{2 sin 30(\cos 2 \theta-1)}{-2 \sin 3 \theta.sin 2\theta}$

$=-\frac{(\cos 2 \theta-1)}{\sin 2 \theta}$

[∵ 1-cos2x=2sin²x

sin 2x=2sinx cosx]

$=\frac{1-\cos 2 \theta}{\sin 2 \theta}$

$=\frac{2 \sin ^2 \theta}{2 \sin \theta \cos \theta}$

=tanθ

(iii) $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$

Sol :

L.H.S

=$\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$

$=\frac{2 \sin \frac{7 x+5 x}{2} \cos 7 \frac{x-5 x}{2}+2 \sin \frac{9 x+3 x}{2} \cos \frac{9 x-3 x}{2}}{2 \cos \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}+2 \cos \frac{9 x+3 x}{2} \cos \frac{9 x-3 x}{2}}$

$=\frac{2 \sin 6 x \cos x+2 \sin 6 x \cos 3 x}{2 \cos 6 x \cos x+2 \cos 6 x \cos 3 x}$

$=\frac{2 \sin 6 x\left(\cos x+\cos 3 x\right)}{2 \cos 6 x(\cos x+\cos 3 x)}$

=tan 6x

(i)
Sol :
$\sin \alpha-\sin \beta=\frac{1}{3}$

$\because \sin C-\sin B=2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$

$2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=\frac{1}{3}$..(i)

$\cos \beta-\cos \alpha=\frac{1}{2}$

$\because \cos C-\cos D=2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$

$2 \sin \frac{\beta+\alpha}{2} \sin\frac{ \alpha-\beta}{2}=\frac{1}{2}$..(ii)

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$\frac{2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{2 \sin \frac{\beta+\gamma}{2} \sin \frac{\alpha-\beta}{2}}=\frac{\frac{1}{3}}{\frac{1}{2}}$

$\cot\frac{\alpha+\beta}{2}=\frac{2}{3}$

(ii)
Sol :

$\operatorname{cosec} A+\sec A=\operatorname{cosec} B+\sec B$

secA-secB=cosecB-cosecA

$\frac{1}{\cos A}-\frac{1}{\cos B}=\frac{1}{\sin B}-\frac{1}{\sin A}$

$\frac{\cos B-\cos A}{\cos A \cos B}=\frac{\sin A-\sin B}{\sin B \sin A}$

$\frac{\sin B \sin A}{\cos A \cos B}=\frac{\sin A-\sin B}{\cos B-\cos A}$

$\tan A \tan B=\frac{2 \cos \frac{A+B}{2}\sin\frac{A-B}{2}}{2 \sin \frac{B+A}{2} \sin \frac{A-B}{2}}$

$\tan A \tan B=\cot \frac{A+B}{2}$

Question 11

Sol :

$\sec (\theta+\alpha)+\sec (\theta-\alpha)=2 \sec \theta$

$\frac{1}{\cos (\theta+\alpha)}+\frac{1}{\cos (\theta-\alpha)}=\frac{2}{\cos \theta}$

$\frac{\cos (\theta-\alpha)+\cos (\theta+\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}=\frac{2}{\cos \theta}$

∵ cos(A+B)+cos(A-B)=2cosA.cosB

∵ cos(A+B).cos(A-B)=cos²A-sin²A

$\frac{2 \cos \theta \cos \alpha}{\cos ^{2} \theta-\sin ^{2} \alpha}=\frac{2}{\cos \theta}$

$\cos ^{2} \theta \cos \alpha=\cos ^{2} \theta-\sin ^{2} \alpha$

$\sin ^{2} \alpha=\cos ^{2} \theta-\cos ^{2} \theta \cos \alpha$

$1^{2}-\cos ^{2} \alpha=\cos ^{2} \theta(1-\cos \alpha)$

$(1-\cos \alpha)(1+\cos \alpha)=\cos ^{2} \theta(1-\cos \alpha)$

$1+\cos \alpha=\cos ^{2} \theta$

Question 12

$\sin 25^{\circ} \cos 115^{\circ}=\frac{1}{2}\left(\sin 140^{\circ}-1\right)$
Sol :
L.H.S
$\sin 25^{\circ} \cos 115^{\circ}$

[∵2sinAcosB=sin(A+B)+sin(A-B)]

$=\frac{1}{2}(2 \sin25^{\circ} \cos 115^{\circ})$

$=\frac{1}{2}[\sin (25^{\circ}+115^{\circ})+\sin (25^{\circ}-115^{\circ})]$

$=\frac{1}{2}[\sin 140^{\circ}+\sin (-90^{\circ})]$

$=\frac{1}{2}[\sin 140^{\circ}-\sin 90^{\circ}]$

$=\frac{1}{2}[\sin 140^{\circ}-1]$

Question 13

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$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{3}{16}$
Sol :
L.H.S
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$

$=\frac{\sqrt{3}}{2} \sin 20^{\circ} \times \frac{1}{2}(2 \sin 80^{\circ} \sin 40^{\circ})$

$=\frac{\sqrt{3}}{4} \sin 20^{\circ}\left(\cos \left(80^{\circ}-40\right)-\cos \left(80^{\circ}+40^{\circ}\right)\right]$

$=\frac{\sqrt{3}}{4} \sin 20^{\circ}\left[\cos 40^{\circ}-\cos 120^{\circ}\right]$

$=\frac{\sqrt{3}}{4} \sin 2 0^{\circ}\left[\cos 40^{\circ}-\cos (180^{\circ}-10^{\circ})\right]$

$=\frac{\sqrt{3}}{4} \sin 20^{\circ}\left[\cos 40^{\circ}-\left(-\cos 60^{\circ}\right)\right]$

$=\frac{\sqrt{3}}{4} \sin 20^{\circ}\left[\cos 40^{\circ}+\frac{1}{2}\right]$

$=\frac{\sqrt{3}}{4} \sin 20^{\circ} \cos 40^{\circ}+\frac{\sqrt{3}}{8} \sin 20^{\circ}$

$=\frac{\sqrt{3}}{8}(2 \sin 20^{\circ} \cos 40^{\circ})+\frac{\sqrt{3}}{8} \sin 20^{\circ}$

$=\frac{\sqrt{3}}{8}[\sin (20^{\circ}+40^{\circ})+\sin (20^{\circ}-40^{\circ})]+\frac{\sqrt{3}}{8} \sin 20^{\circ}$

$=\frac{\sqrt{3}}{8}\left(\sin 60^{\circ}+\sin \left(-20^{\circ}\right)\right]+\frac{\sqrt{3}}{8} \sin 20^{\circ}$

$=\frac{\sqrt{3}}{8}\left[\frac{\sqrt{3}}{2}-\sin 20^{\circ}\right]+\frac{\sqrt{3}}{8} \sin 20^{\circ}$

$=\frac{3}{16}-\frac{\sqrt{3}}{8} \sin 20^{\circ}+\frac{\sqrt{3}}{8} \sin 20^{\circ}$

$=\frac{3}{16}$

Question 14

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$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{16^{\circ}}$
Sol :
L.H.S

$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$

$=\frac{1}{2} \cos 20^{\circ} \times \frac{1}{2}(2 \cos 80^{\circ} \cos 40^{\circ})$

[∵ 2cosAcosB=cos(A+B)+cos(A-B)]

$=\frac{1}{4} \cos 20^{\circ}[\cos (80^{\circ}+40^{\circ})+\cos (80^{\circ}-40^{\circ})]$

$=\frac{1}{4} \cos 20^{\circ}\left[\cos 120^{\circ}+\cos 40^{\circ}\right]$

$=\frac{1}{4} \cos 20^{\circ}\left[\cos (180^{\circ}-60^{\circ})+\cos 40^{\circ}\right]$

$=\frac{1}{4} \cos 20^{\circ}[-\cos 60^{\circ}+\cos 40^{\circ}]$

$=\frac{1}{4} \cos 20^{\circ}\left[-\frac{1}{2}+\cos 40^{\circ}\right]$

$=-\frac{1}{8} \cos 20^{\circ}+\frac{1}{4} \cos 40^{\circ} \cos 20^{\circ}$

$=-\frac{1}{8} \cos 20^{\circ}+\frac{1}{8}(2 \cos 40^{\circ} \cos 20^{\circ})$

$=-\frac{1}{8} \cos 20^{\circ}+\frac{1}{8}[\cos (40^{\circ}+20^{\circ})+\cos (40^{\circ}-20^{\circ})]$

$=-\frac{1}{8} \cos 20^{\circ}+\frac{1}{8}[\cos 60^{\circ}+\cos 20^{\circ}]$

$=-\frac{1}{8} \cos 20^{\circ}+\frac{1}{8} \cos 60^{\circ}+\frac{1}{8} \cos 20^{\circ}$

$=\frac{1}{8} \times \frac{1}{2}$

$=\frac{1}{16}$

Question 15

Question 17

$4 \cos \theta \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)=\cos 3 \theta$
Sol :
L.H.S
$4 \cos \theta \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)$

$=2 \cos \theta \cdot 2 \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)$

[∵ 2cosA.cosB=cos(A+B)+cos(A-B)]

$=2 \cos \theta\left[\cos \left\{\left(\frac{\pi}{3}+\theta\right)+\left(\frac{\pi}{3}-\theta\right)\right\}\right.+\cos \left\{\left(\frac{\pi}{3}+\theta\right)-\left(\frac{\pi}{2}-\theta \right)\right]$

$=2 \cos \theta\left[\cos \left(\frac{\pi}{3}+\theta+\frac{\pi}{3}-\theta\right)+\cos \left(\frac{\pi}{3}+\theta-\frac{\pi}{3}+\theta\right)\right.$

$=2 \cos \theta\left[\cos \frac{2 \pi}{3} +\cos 2 \theta\right]$

$=2 \cos \theta\left[\cos \left(\pi-\frac{\pi}{3}\right)+\cos 2 \theta\right]$

$=2 \cos \theta\left[-\cos \frac{\pi}{3}+\cos 2 \theta\right]$

$=2 \cos \theta\left[-\frac{1}{2}+\cos 2 \theta\right]$

$=-\cos \theta+2 \cos 2 \theta \cos \theta$

$=-cos \theta+\cos (2 \theta+\theta)+\cos (2 \theta-\theta)$

$=-\cos \theta+\cos 3 \theta+\cos \theta$

Question 18

$\tan \theta \tan \left(60^{\circ}+\theta\right) \tan \left(60^{\circ}-\theta\right)=\tan 3 \theta$
Sol :
L.H.S
$\tan \theta \tan (60+\theta) \tan (60-\theta)$

$=\frac{\sin \theta \sin (60^{\circ}+\theta) \sin (60^{\circ}-\theta)}{\cos \theta \cos (60^{\circ}+\theta) \cos (60^{\circ}-\theta)}$

[∵ sin(A+B)sin(A-B=sin²A.sin²B]

[∵ cos(A+B)cos(A-B)=cos²B-sin²A]

$=\frac{\sin \theta\left[\sin ^{2} 60^{\circ}-\sin ^{2} \theta\right]}{\cos \theta\left[\cos ^{2} \theta-\sin ^{2} 60^{\circ}\right]}$

$=\frac{\sin \theta\left[\left(\frac{\sqrt{3}}{2}\right)^{2}-\sin ^{2} \theta\right]}{\cos \theta\left[\cos ^{2} \theta-\left(\frac{\sqrt{3}}{2}\right)^{2}\right]}$

$=\frac{\sin \theta\left[\frac{3}{4}-\sin ^{2} \theta\right]}{\cos \theta\left[\cos ^{2} \theta-\frac{3}{4}\right]}$

$=\frac{\sin \theta\left[\frac{3-4 \sin ^{2} \theta}{4}\right]}{\cos \theta\left[\frac{4 \cos ^{2} \theta-3}{4}\right]}$

$=\frac{3 \sin \theta-4 \sin 3 \theta}{4 \cos ^{3} \theta-3 \cos \theta}$

$=\frac{\sin 3 \theta}{\cos 3 \theta}=\tan 3 \theta$

Question 19

Sol :
$\cos \alpha \cos \beta=\frac{1}{2} \times 2 \cos \alpha \cos \beta$

$\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]$

$\cos \alpha \cos \beta=\frac{1}{2}[\cos 90+\cos (\alpha-\beta)]$

$\cos \alpha \cos \beta=\frac{1}{2} \cos (\alpha-\beta)$

$(\cos (\alpha-\beta) \leq 1]$

∴ $\cos \alpha \cos \beta \leq \frac{1}{2}(1)$

$\cos \alpha \cos \beta \leq \frac{1}{2}$

Question 20

Sol :

$\cos \alpha=\frac{1}{\sqrt{2}}$

[]

$\sin \alpha=\pm \sqrt{1-\cos ^{2} \alpha}$

$=\pm \sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}$

$=\pm \sqrt{1-\frac{1}{2}}$

$=\pm \sqrt{\frac{2-1}{2}}$

$=\pm \frac{1}{\sqrt{2}}$

$\sin \alpha=+\frac{1}{\sqrt{2}}$ , $\sin \alpha=-\frac{1}{\sqrt{2}}$

L.H.S

$\tan \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$

$=\frac{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}{2 cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}$

$=\frac{\sin \alpha+\sin \beta}{\sin \alpha-\sin \beta}$

CASE-I
$\sin \alpha=\frac{1}{\sqrt{2}}, \quad \sin \beta=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{\sin \alpha+\sin A}{\sin \alpha-\sin \beta}$

$=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}}$

$=\frac{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}}{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

[]

$=\frac{3+\sqrt{6}+\sqrt{6}+2}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$=\frac{5+2 \sqrt{6}}{3-2}$

$=5+2 \sqrt{6}$

CASE-II

$\sin \alpha=-\frac{1}{\sqrt{2}} \quad, \sin \beta=\frac{1}{\sqrt{3}}$

$\frac{\sin \alpha+\sin \beta}{\sin \alpha-\sin \beta}=5-2 \sqrt{6}$

Question 21

Sol :
$x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)=k$

$\cos \theta=\frac{k}{x}$  , $\cos \left(\theta+\frac{2 \pi}{3}\right)=\frac{k}{y}$ , $\cos \left(\theta+\frac{4 \pi}{3}\right)=\frac{k}{z}$

[]

$\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right)=\frac{k}{x}+\frac{k}{y}+\frac{k}{z}$

$\cos \theta+\cos \left[\pi-\left(\frac{\pi}{3}-\theta\right)\right]+\cos \left[\pi+\left(\frac{\pi}{3}+\theta\right)\right]=k\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$

$\cos \theta-\cos \left(\frac{\pi}{3}-\theta\right)-\cos \left(\frac{\pi}{3}+\theta\right)=k\left(\frac{y z+x z+x y}{x y z}\right)$

$\cos \theta-\left[\cos \left(\frac{\pi}{3}-\theta\right)+\cos \left(\frac{\pi}{3}+\theta\right)\right]=k \frac{\left(xy+yz+zx\right)}{xyz}$

$\cos \theta-\left[2 \cos \frac{\pi}{3} \cos \theta\right]= \frac{k(x y+y z+z x)}{x y z}$

$\cos \theta-\left[2 \times \frac{1}{2} \cos \theta\right]=\frac{k(x y+yz+z x)}{x y z}$

$0=k\left(\frac{x y+y z+z x}{x y z}\right)$

0=xy+yz+zx

Question 22

Sol :
$y \sin \phi=x \sin (2 \theta+\phi)$

$\frac{y}{x}=\frac{\sin (2 \theta+\phi)}{\sin \phi}$

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$\frac{y+x}{y-x}=\frac{\sin (2 \theta+\phi)+\sin \phi}{\sin (2 \theta+\phi)-\sin \phi}$

$\frac{x+y}{y-x}=\frac{2 \sin \frac{2 \theta+\phi+\phi}{2} \cos \frac{2 \theta+\phi-\phi}{2}}{2 \cos \frac{2 \theta+\phi+\phi}{2} \sin \frac{2 \theta+\phi-\phi}{2}}$

$\frac{x+y}{y-x}=\frac{\sin \frac{2(\theta+\phi)}{2}\cos \frac{2\theta}{2}}{\cos \frac{2(\theta+\phi)}{2} \sin \frac{2 \theta}{2}}$

$\frac{x+y}{y-x}=\tan (\theta+\phi) \cot \theta$

$\frac{x+y}{\tan (\theta+\phi)}=(y-x) \cot \theta$

$(x+y) \cot (\theta+\phi)=(y-x) \cot \theta$

Question 23

Sol :
$\cos (\alpha+\beta) \sin (\gamma+\delta)=\cos (\alpha-\beta) \sin (\gamma-\delta)$

$\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{\sin (\gamma-\delta)}{\sin (\gamma+\delta)}$

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$\frac{\cos (\alpha+\beta)+\cos (\alpha-\beta)}{\cos (\alpha+\beta)-\cos (\alpha-\beta)}=\frac{\sin (\gamma-\delta)+\sin (\gamma+\delta)}{\sin (\gamma-\delta)-\sin (\gamma+\delta)}$

$\frac{2 \cos \alpha \cos \beta}{-2 \sin \alpha \sin \beta}=\frac{2 \sin \gamma \cos \delta}{-2 \cos \gamma \sin \delta}$

$\cot \alpha \cot \beta=\tan \gamma \cot \delta$

$\frac{\cot \alpha \cos \beta}{\tan \gamma}=\cos \delta$

$\cot \alpha \cot \beta \cot \gamma=\cot\delta$

Question 24

Sol :
$\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=0$

$\frac{\cos (A-B)}{\cos (A+B)}=\frac{-\cos (C+D)}{\cos (C-D)}$

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$\frac{\cos (A-B)+\cos (A+B)}{\cos (A-B)-\cos (A+B)}=\frac{-\cos (C+D)+\cos (C-D)}{-cos(C+D)-\cos (C-D)}$

$\frac{2 \cos A \cos B}{2 \sin A \sin B}=\frac{2 \sin C \sin D}{-2 \cos C \cos D}$

cotA.cotB=-tanC.tanD

$-1=\frac{\tan C \tan B}{\cot A \cot B}$

-1=tanA.tanB.tanC.tanD

Question 25

Sol :
(i)
$\tan (\theta+\phi)=3 \tan \theta$

$\frac{\tan (\theta+\phi)}{\tan \theta}=\frac{3}{1}$

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$\frac{\tan (\theta+\phi)-\tan \theta}{\tan (\theta+\phi)+\tan \theta}=\frac{3-1}{3+1}$

$\dfrac{\frac{\sin (\theta+\phi)}{\cos (\theta+\phi)}-\frac{\sin \theta}{\cos \theta}}{\frac{\sin (\theta+\phi)}{\cos (\theta+\phi)}+\frac{\sin \theta}{\cos \theta}}=\frac{2}{4}$

$\dfrac{ \frac{\sin(\theta+\phi) \cos \theta-\cos (\theta+\phi) \sin \theta}{\cos (\theta+\phi) \cos \theta}} {\frac{\sin (\theta+\phi) \cos \theta+\cos (\theta+\phi) \sin \theta}{{\cos(\theta+\phi) \cos \theta}}}=\frac{1}{2}$

$\dfrac{\sin \left[\theta+\phi-\theta\right]}{\sin [\theta+\phi+\theta]}=\frac{1}{2}$

$\frac{\sin \phi}{\sin (2 \theta+\phi)}=\frac{1}{2}$

$\sin (2 \theta+\phi)=2 \sin \phi$

(ii)
$\sin (2 \theta+\phi)=2 \sin \phi$

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[∵ sin2A=2sinA.cosA]

$\sin (2 \theta+\phi) \cdot \cos \phi=2 sin \phi cos\phi$

$\frac{1}{2} \times[2 \sin (2 \theta+\phi) \cos \phi]=\sin 2 \phi$

$\frac{1}{2}[\sin (2 \theta+\phi+\phi)+\sin (2 \theta+\phi-\phi)]=\sin 2 \phi$

$\sin (2 \theta+2 \phi)+\sin 2 \theta=2 \sin 2 \phi$

$\sin 2(\theta+\phi)+\sin 2 \theta=2 \sin 2 \phi$