KC Sinha Mathematics Solution Class 11 Chapter 6 त्रिकोणमितीय फलन (Trigonometric function) Exercise 6.3

Exercise 6.3

Question 1

Sol :
$\because \sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}$

$\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$=\frac{2\left(\frac{a}{b}\right)}{1+\left(\frac{a}{b}\right)^{2}}$

$=\frac{\frac{2 a}{b}}{\frac{b^{2}+a^{2}}{b^{2}}}$

$=\frac{2 a}{b} \times \frac{b^{2}}{b^{2}+a^{2}}$

$\sin 2 \theta=\frac{2 a b}{a^{2}+b^{2}}$


$\because \cos 2 A=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} A}$

$\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\frac{1-\left(\frac{a}{b}\right)^{2}}{1+\left(\frac{a}{b}\right)^{2}}$

$=\frac{1-\frac{a^{2}}{b^{2}}}{1+\frac{a^{2}}{b^{2}}}$

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

$\cos 2 \theta=\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$


$\tan 2 A=\frac{2 \tan A}{1-\tan ^{2}}$

$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$

$=\frac{2\left(\frac{a}{b}\right)}{1-\left(\frac{a}{b}\right)^{2}}$

$=\frac{\frac{2 a}{b}}{1-\frac{a^{2}}{b^{2}}}$

$=\frac{\frac{2 q}{b}}{\frac{b^{2}-a^{2}}{b^{2}}}$

$\tan 2 \theta=\frac{2 a}{b} \times \frac{b^{2}}{b^{2}-a^{2}}$

$\tan 2 \theta=\frac{2 a b}{b^{2}-a^{2}}$

Question 2

Sol :
(i)
$\tan \theta=-\frac{3}{4}, \quad \frac{\pi}{2}<\theta<\pi$

Question 3

cosθ-tanθ=2cot2θ
Sol :
L.H.S
cosθ-tanθ

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

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

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

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

$=2 \cot 2 \theta$

Question 4

$\frac{1-\tan ^{2}\left(\frac{\pi}{4}-A\right)}{1+\tan ^{2}\left(\frac{\pi}{4}-A\right)}=\sin 2 A$
Sol :
L.H.S

$\because \cos A=\frac{1-\tan ^{2} A}{1+\tan ^{2} B}$

$\frac{1-\tan ^{2}\left(\frac{\pi}{4}-A\right)}{1+\tan ^{2}\left(\frac{\pi}{4}-A\right)}$

$=\cos 2\left(\frac{\pi}{4}-A\right)$

$=\cos \left(\frac{\pi}{2}-2A\right)=\sin 2A$

Question 5

$\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\sec 2 \theta-\tan 2 \theta$
Sol :
R.H.S

sec2θ-tan2θ

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

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

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

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

$=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}$

Question 6

$\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right)=2 \tan 2 \theta$
Sol :
$\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right)$

$=\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4}\tan \theta}-\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}$

$=\frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta}$

$=\frac{(1+\tan \theta)^{2}-(1-\tan \theta)^{2}}{(1-\tan \theta)(1+\tan \theta)}$

$=\frac{4 \times{1} . \tan \theta}{1^{2}-\tan ^{2} \theta}$

$=\frac{2 \cdot 2 \tan \theta}{1-\tan ^{2} \theta}$

$=\left[\tan 2 A=\frac{2 \tan \theta}{-\tan ^{2} \theta}\right]$

=2tan2θ

Question 7

Sol :
$\cos 2 \theta=2 \cos ^{2} \theta-1$

$=2\left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]^{2}-1$

$=2 \times \frac{1}{4}\left(a^{2}+\frac{1}{a^{2}}+2 \cdot a \cdot \frac{1}{a}\right)-1$

$=\frac{1}{2}\left(a^{2}+\frac{1}{a^{2}}+2\right)-1$

$=\frac{1}{2}\left(a^{2}+\frac{1}{a^{2}}\right)+1-1$

⇒$\cos 2 \theta=\frac{1}{2}\left(a^{2}+\frac{1}{a^{2}}\right)$

Question 8

$\cos ^{2} \theta+\sin ^{2} \theta \cos 2 \beta=\cos ^{2} \beta+\sin ^{2} \beta \cos 2 \theta$
Sol :
L.H.S
$\cos ^{2} \theta+\sin ^{2} \theta \cos 2 \beta$

$=\cos ^{2} \theta+\sin ^{2} \theta\left(1-2 \sin ^{2} \beta\right)$

$=\cos^{2} \theta+\sin ^{2} \theta-2 \sin ^{2} \theta \sin ^{2} \beta$

$=1-2 \sin ^{2} \theta \sin ^{2} \beta$

$=\cos^{2} \beta+\sin ^{2} \beta-2 \sin ^{2} \theta \sin ^{2} \beta$

$=\cos^{2} \beta+\sin ^{2} \beta\left(1-2 \sin ^{2} \theta\right)$

$=\cos ^{2} \beta+\sin ^{2} \beta \cos 2 \theta$

Question 9

1+tanθ.tan2θ=sec2θ
Sol :
1+tanθ.tan2θ

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

$=\frac{\cos \theta \cos 2 \theta+ \sin \theta \sin2 \theta}{\cos \theta \cos 2 \theta}$

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

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

$=\dfrac{\cos\theta}{\cos\theta\cos 2\theta}$

Question 10

$\frac{1+\sin 2 A-\cos 2 A}{1+\sin 2 A+\cos 2 A}=\tan A$
Sol :
L.H.S

$\frac{1+\sin 2 A-\cos 2 A}{1+\sin 2 A+\cos 2 A}$

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

$=\frac{2 \sin A \cos A+2 \sin ^{2} A}{2 \sin A \cos A+2 \cos^2 A}$

$=\frac{2 \sin A\left(cosA+\sin A\right)}{2 \cos A(\sin A+\cos A)}$

Question 11

$\frac{1+\sin 2 \theta}{1-\sin 2 \theta}=\left(\frac{1+\tan \theta}{1-\tan \theta}\right)^{2}$
Sol :
L.H.S

$\frac{1+\sin 2 \theta}{1-\sin 2 \theta}$

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

∵1+sin2A=(cosA+sinA)²

∵1+sin2A=(cosA-sinA)²

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

$=\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)^{2}$

$=\left(\frac{\frac{\cos \theta+\sin \theta}{\cos \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}\right)^{2}$

$=\left(\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}\right)^{2}$

$=\left(\frac{1+\tan \theta}{1-\tan \theta}\right)^{2}$

Question 12

$\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=4$
Sol :
L.H.S

$\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$

$\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$

$=\frac{1 \cdot \cos 10^{\circ}-\sqrt{3} \cdot \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$

$=\frac{2\left(\frac{1 \cdot \cos 10^{\circ}-\sqrt{3} \cdot \sin 10^{\circ}}{2}\right)}{\sin 10^{\circ} \cos10^{\circ}}$

$=\frac{2\left(\frac{1 \cdot \cos 10^{\circ}-\sqrt{3} \cdot \sin 10^{\circ}}{2}\right)}{\sin 10^{\circ} \cos10^{\circ}}$

$=2 \frac{(\sin 30^{\circ} \cos 10^{\circ}-\cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ}{\cos 10^{\circ}}}$

$=\frac{4 \cdot \sin \left(30^{\circ}-10^{\circ}\right)}{2 \sin 10^{\circ} \cos 10^{\circ}}$

$=\frac{4 \sin 20^{\circ}}{\sin 2(10^{\circ})}=\frac{4 \sin 2 0^{\circ}}{\sin 20^{\circ}}=4$

Question 13

cosecA-2cot2A.cosA=2sinA
Sol :
L.H.S

cosecA-2cot2AcosA

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

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

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

$=\frac{2 \sin ^{2} A}{\sin A}$

=2sinA

Question 14

$\cot ^{2} A-\tan ^{2} A=4 \cot 2 A \operatorname{cosec} 2 A$
Sol :
$\cot^{2} A-\tan ^{2} A$

$=\frac{\cos^{2} A}{\sin ^{2} A}-\frac{\sin^{2} A}{\cos ^{2} A}$

$=\frac{\cos 4 A-\sin 4 A}{\sin ^{2} A \cos ^{2} A}$

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

$=\frac{\left(\cos ^{2} A-\sin ^{2} A\right)\left(\cos^{2} A+\sin ^{2} A\right)}{\sin ^{2} A \cos ^{2} A}$

$=\frac{\cos ^{2} A-\sin^{2} A}{\sin ^{2} A \cos ^{2} A}$

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

$=\frac{4 \cos 2 A}{4 \sin ^{2} A \cos ^{2} A}$

$=\frac{4 \cos 2 \theta}{(2 \sin \theta \cos \theta)^{2}}$

$=\frac{4 \cos 2 A}{\sin ^{2} 2 \theta}$

$=\frac{4 \cos 2 A}{\sin 2 A} \times \frac{1}{\sin 2 A}$

=4cot2A.cosec2A

Question 15

$\frac{1+\sin 2 A}{\cos 2 A}=\frac{\cos A+\sin A}{\cos A-\sin A}=\tan \left(\frac{\pi}{4}+A\right)$
Sol :
$\frac{1+\sin 2 A}{\cos 2 A}=\frac{(\cos A+\sin A)^{2}}{\cos ^{2} A-\sin ^{2} A}$

$=\frac{(\cos A+\sin A)^{2}}{(\cos A-\sin A)\left(\cos A+\sin A\right)}$

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

[]

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

$=\frac{\tan \frac{\pi}{4}+\tan A}{1-\tan \frac{\pi}{4} \cdot \tan A}$

$=\tan \left(\frac{\pi}{4}+A\right)$

Question 16 

$\cos ^{6} A-\sin ^{6} A=\cos 2 A\left(1-\frac{1}{4} \sin ^{2} 2 A\right)$

Sol: 

L.H.S

$\cos ^{6} A-\sin ^{6} A=\left(\cos ^{2} A\right)^{3}-\left(\sin ^{2} A \right)^{3}$

$=\left[\cos ^{2} A-\sin ^{2} A\right)\left[\left(\cos^{2} A\right)^{2}+\cos ^{2} A \sin ^{2} A+(sin^{2}A)^{2} \right]$

$=\cos 2 A\left[\left(cos^{2} A\right)^{2}+\left(\sin ^{2} A\right)^{2}+\cos ^{2} A \sin ^{2} \theta\right]$

$=\cos 2 A\left[\left(\cos^{2} A+\sin ^{2} A\right)^{2}-2 \cos ^{2} A \sin ^{2} A+\cos^{2}A\sin^{2}A\right]$

$=\cos 2 A\left[1-\cos ^{2} A \sin ^{2} A\right]$

$=\cos 2 A\left[1-\frac{1}{4} \times 4 \cos ^{2} A \sin ^{2} A\right]$

$=\cos 2 \alpha\left[1-\frac{1}{4}(2 \cos A \sin A)^{2}\right]$

$=\cos A\left[1-\frac{1}{4} \sin ^{2} 2 A\right]$

Question 17

$\cos ^{2} \theta+\cos ^{2}\left(\frac{\pi}{3}+\theta\right)+\cos ^{2}\left(\frac{\pi}{3}-\theta\right)=\frac{3}{2}$
Sol :
L.H.S

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

$=\frac{1}{2}\left(2 \cos ^{2} \theta\right)+\frac{1}{2} \cdot 2 \cos ^{2}\left(\frac{\pi}{3}+\theta\right)+\frac{1}{2} \cdot 2 \cos ^{2}\left(\frac{\pi}{3}-\theta\right)$

$=\frac{1}{2}[1+\cos 2 \theta]+\frac{1}{2}\left[1+\cos 2\left(\frac{\pi}{3}+\theta\right)\right]+\frac{1}{2}\left[1+\cos 2\left(\frac{\pi}{3}-\theta\right)\right.$

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

$=\frac{3}{2}+\frac{1}{2} \cos 2 \theta+\frac{1}{2}\left[\cos \left(\frac{2 \pi}{3}+2 \theta\right)+\cos \left(\frac{2 \pi}{3}-2 \theta\right)\right]$

$=\frac{3}{2}+\frac{1}{2} \cos 2 \theta+\frac{1}{2} \times 2 \cos \frac{2 \pi}{3} \cos 2 \theta$

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

$=\frac{3}{2}+\frac{1}{2} \cos 2 \theta-\cos \frac{\pi}{3} \cos 2 \theta$

$=\frac{3}{2}+\frac{1}{2} \cos 2 \theta-\frac{1}{2} \cos {2} \theta$

$=\frac{3}{2}$

Question 18

$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)=13$
Sol :
L.H.S

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

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

$=3[1-\sin 2 x]^{2}+6(1+\sin 2 x)+4\left[\left(\sin ^{2} x+\cos ^{2} x\right)\right. \left\{\left(\sin ^{2} x\right)^{2}\right.\left.-\sin ^{2} x \cdot \cos ^{2} x+\left(\cos ^{2} x\right)^{2}\right]$

$=3\left[1^{2}-2 \cdot 1 \cdot \sin 2 x+\sin ^{2} 2 x\right]+6(1+\sin 2 x)+4[(1) \left[\left(\sin^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}-\sin ^{2} x \cos ^{2} x\right]$

$=3\left[1-2 \sin 2 x+\sin ^{2} 2 x\right]+6(1+\sin 2 x) \left.+4\left[\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x\right]$

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

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

$=13+3(2 \sin x \cos x)^{2}-12 \sin ^{2} 2 \cos ^{2} x$

$=13+12 \sin ^{2} x \cos ^{2} x-12 \sin ^{2} x \cos ^{2} x$

Question 19

$2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1=0$
Sol :
L.H.S
$2\left(\sin^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$

$=2\left[\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right]-3\left[\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}\right]+1$

$\begin{aligned}=2\left[\left(\sin ^{2} x+\cos ^{2} x\right)\left[\left(\sin ^{2} x\right)^{2}-\sin ^{2} x \cdot \cos ^{2} x+\left(\cos ^{2} x\right)^{2}\right\}\right] -3\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right] & \end{aligned}$

$=2\left[\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}-\sin ^{2} x \cos ^{2} x\right]-3\left[1-2 \sin ^{2} x \cos ^{2} x\right]+1$

$=2\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2}x-\sin ^{2}x \cos ^{2} x\right]-3\left[1-2 \sin ^{2} x \cos ^{2}x\right]+1$

$=2\left[1-3 \sin ^{2} x \cos^{2} x\right]-3\left[1-2 \sin ^{2} x \cos ^{2} 2\right]+1$

$=2-6 \sin ^{2} x \cos ^{2} x-3+6 \sin ^{2}x \cos ^{2} x+1$

Question 20

Sol :
$\cos ^{2} \theta+\cos ^{2}(\alpha+\theta)-2 \cos \alpha \cos \theta \cos (\alpha+\theta)$

$=\cos ^{2} \theta+\cos ^{2}(\alpha+\theta)-[\cos (\alpha+\theta)+\cos (\alpha-\theta)] \cos \left(\alpha+\beta\right)]$

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

$=\cos ^{2} \theta-\left(\cos ^{2} \theta-\sin ^{2} \alpha\right)$

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

Question 21

$4\left(\cos ^{3} 10^{\circ}+\sin ^{3} 20^{\circ}\right)=3\left(\cos 10^{\circ}+\sin 20^{\circ}\right)$
Sol :
L.H.S
$4\left(\cos ^{3} 10^{\circ}+\sin ^{3} 20^{\circ}\right)$

$=4 \cos ^{3} 10^{\circ}+4 \sin ^{3} 20^{\circ}$

$\because 4 \sin ^{3} A=3 \sin A-\sin 3A$

$\because 4 \cos ^{3} A=cos3A+3cosA$

$=\cos 3(10^{\circ})+3 \cos 10^{\circ}+3 \sin 20^{\circ}-\sin 3\left(20^{\circ}\right)$

$\cos 30^{\circ}+3 \cos 10^{\circ}+3 \sin 20^{\circ}-\sin 60^{\circ}$

$=\frac{\sqrt{3}}{2}+3\left(cos10^{\circ}+\sin 20^{\circ}\right)-\frac{\sqrt{3}}{2}$

$=3(\cos 10^{\circ}+\sin 20^{\circ})$

Question 22

$\sin \theta \cos ^{3} \theta-\cos \theta \sin ^{3} \theta=\frac{1}{4} \sin 4 \theta$
Sol :
L.H.S

$\sin \theta \cos ^{3} \theta-\cos \theta \sin ^{3} \theta$

$=\sin \theta \times \frac{1}{4} \times 4 \cos ^{3} \theta-\cos \theta \frac{1}{4} \times 4 \sin ^{3} \theta$

$=\frac{1}{4} \sin \theta(\cos 3 \theta+3 \cos \theta)-\frac{1}{4} \cos \theta(3 \sin \theta-\sin3 \theta)$

$\begin{aligned}=\frac{1}{4} \sin \theta \cos 30+\frac{3}{4} \sin \theta \cos \theta &-\frac{3}{4} \sin \theta \cos \theta +\frac{1}{4} \cos \theta \sin \frac{3 \theta}{2} \end{aligned}$

$=\frac{1}{4}[\sin \theta \cos 3 \theta+\cos \theta \sin 3 \theta]$

$=\frac{1}{4} \sin(\theta+3 \theta)$

$=\frac{1}{4} \sin 4 \theta$

Question 24

$\tan \theta+\tan \left(60^{\circ}+\theta\right)+\tan \left(120^{\circ}+\theta\right)=3 \tan 3 \theta$
Sol :
L.H.S
$\tan \theta+\tan (60^{\circ}+\theta)+\tan (120^{\circ}+\theta)$

$=\tan \theta+\tan (60+\theta)+\tan [180^{\circ}-(60^{\circ}-\theta)]$

$=\tan \theta+\tan (60+\theta)+\tan [(60^{\circ}-\theta)]$

$=\tan \theta+\frac{\tan 60^{\circ}+\tan \theta}{1-\tan 60^{\circ} \tan \theta}-\frac{\tan 60^{\circ}-\tan \theta}{1+\tan 60^{\circ} \cdot \tan \theta}$

$=\tan \theta+\frac{\sqrt{3}+\tan \theta}{1-\sqrt{3} \tan \theta}-\frac{\sqrt{3}-\tan \theta}{1+\sqrt{3} \tan \theta}$

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

$=\tan \theta+\frac{\left(\sqrt{3}+3 \tan \theta+\tan \theta+\sqrt{3} \tan ^{2} \theta\right)-(\sqrt{3}-\tan \theta -3\tan \theta+\sqrt{3}tan^{2}\theta)}{(1)^{2}-(\sqrt{3} \tan ^{2} \theta)^{2}}$

$=\tan \theta +\dfrac{(\sqrt{3}+4\tan \theta +\sqrt{3} \tan^{2} \theta)-(\sqrt{3}-4 \tan \theta +\sqrt{3} \tan^{2}\theta)}{1-3\tan^{2} \theta}$

$=\tan \theta+\dfrac{\sqrt{3}+4 \tan \theta+\sqrt{3} \tan ^{2} \theta-\sqrt{3}+4 \tan \theta-\sqrt{3} \tan ^{2} \theta}{ 1-3 \tan ^{2} \theta}$

$=\frac{\tan \theta}{1}+\frac{8 \tan \theta}{1-3 \tan ^{2} \theta}$

$=\frac{\tan \theta\left(1-3 \tan ^{2} \theta\right)+8 \tan \theta}{1-3 \tan ^{2} \theta}$

$=\frac{\tan \theta-3 \tan ^{3} \theta+8 \tan \theta}{1-3 \tan ^{2} \theta}$

$=\frac{9 \tan \theta-3 \tan ^{3} \theta}{1-3 \tan ^{2} \theta}$

$=\frac{3\left(3 \tan \theta-\tan ^{3} \theta\right)}{1-3 \tan ^{2} \theta}$

$=3 \tan 3\theta$

Question 25

$4 \sin \theta \sin \left(\theta+\frac{\pi}{3}\right) \sin \left(\theta+\frac{2 \pi}{3}\right)=\sin 3 \theta$
Sol :
L.H.S
$4 \sin \theta \sin \left(\theta+\frac{\pi}{3}\right) \sin \left(\theta+\frac{2 \pi}{3}\right)$

$=4 \sin \theta \sin \left(\frac{\pi}{3}+\theta\right) \cdot \sin \left[\pi-\left(\frac{\pi}{3}-\theta\right)\right]$

$=4 \sin \theta \sin \left(\frac{\pi}{3}+\theta\right) \cdot \sin \left(\frac{\pi}{3}-\theta\right)$

$=\sin (A+B) \sin \left(A{-} B\right) =\sin ^{2} A{-\sin ^{2}B}$

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

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

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

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

$=\sin 3 \theta$

Question 27

(i) $\sin 5 \theta=5 \sin \theta-20 \sin ^{3} \theta+16 \sin ^{5} \theta$
Sol :
L.H.S
$\sin 5 \theta=\sin (3 \theta+2 \theta)$

$=\sin3\theta .\cos 2\theta+\cos3\theta.\sin2\theta$

$=\left(3 \sin \theta-4 \sin ^{3} \theta\right)\left(1-2 \sin ^{2} \theta\right)+\left(4 \cos ^{3} \theta-3 \cos \theta\right) 2 \sin \theta \cos \theta$

$=3 \sin \theta-6 \sin ^{3} \theta-4 \sin ^{3} \theta+8 \sin ^{5} \theta+\left(4 \cos ^{2} \theta-3\right) \cdot 2 \sin \theta \cos ^{2} \theta$

$=3 \sin \theta-10 \sin ^{3} \theta+8 \sin ^{5} \theta+\left[4\left(1-\sin ^{2} \theta\right)-3\right]. 2 \sin \theta .(1-sin^{2} \theta)$

$=3 \sin \theta-10 \sin ^{3} \theta+8 \sin ^{5} \theta+\left(4-4 \sin ^{2} \theta-3\right)\left(2 \sin \theta-2 \sin^{3} \theta\right)$

$=3 \sin \theta-10 \sin ^{3} \theta+8 \sin ^{5} \theta+\left(1-4 \sin ^{2} \theta\right)\left(2 \sin \theta-2 \sin ^{3} \theta\right)$

$=3 \sin \theta-10 \sin ^{3} \theta+8 \sin ^{5} \theta+2 \sin \theta-2 \sin ^{3} \theta-8 \sin ^{3} \theta+8sin^{5} \theta$

$=5 \sin \theta-20 \sin ^{3} \theta+16 \sin ^{5} \theta$



(ii) $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$
Sol :
R.H.S
$1-8 \sin ^{2} x \cos ^{2} x$

$=1-2 \cdot\left(4 \sin ^{2} x \cos ^{2} x\right)$

[∵ sin2A=2sinA.cosA]

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

$=1-2 \sin ^{2} 2 x$

$\because \cos2 A=1-2 \sin ^{2}A$

=cos2(2x)
=cos4x

Question 28

$\cos 6 \theta=32 \cos ^{6} \theta-48 \cos ^{4} \theta+18 \cos ^{2} \theta-1$
Sol :
L.H.S
$\cos 6 \theta=\cos 2(3 \theta)$

$\because \cos2A=2\cos^{2}A-1$

$=2 \cos ^{2} 3 \theta-1$

$=2\left[4 \cos ^{3} \theta-3 \cos \theta\right]^{2}-1$

$=2\left[(4 \cos 3 \theta]^{2}-2 \cdot 4 \cos ^{3} \theta \cdot 3 \cos \theta+(3 \cos \theta)^{2}-1\right)$

$=2[16\cos^{6} \theta-24\cos^{4} \theta+9\cos^{2}\theta]-1$

$=32\cos^ 6 \theta-48 \cos^{4} \theta+18\cos ^{2} \theta-1$

Question 29

$\cos 4 \theta-\cos 4 \alpha=8(\cos \theta-\cos \alpha)(\cos \theta+\cos \alpha)(\cos \theta-\sin \alpha)(\cos \theta+\sin \alpha)$
Sol :
L.H.S

$\cos 4 \theta-\cos 4\alpha$

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

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

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

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

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

$=2\left[\left(2 \cos ^{2} \theta-1\right)-\left(2 \cos ^{2} \alpha-1\right)\right]\left[\left(2 \cos ^{2} \theta-1\right)+\left(1-2 \sin ^{2} \alpha\right)\right]$

$=2\left[2 \cos ^{2} \theta-1-2 \cos ^{2} \alpha+1\right]\left[2 \cos ^{2} \theta-1+1-2 \sin ^{2} \alpha\right]$

$=8\left(\cos ^{2} \theta-\cos ^{2} \alpha\right)\left(\cos ^{2} \theta-\sin ^{2} \alpha\right)$

$=8(\cos \theta-\cos \alpha)(\cos \theta+\cos \alpha)(\cos \theta-\sin \alpha)(\cos \theta+\sin \alpha)$

Question 30

Sol :
L.H.S

acos2x+bsin2x

$=a\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)+b\left(\frac{2 \tan x}{1+\tan ^{2} x}\right)$

$=a\left[\frac{1-\left(\frac{b}{a}\right)^{2}}{1+\left(\frac{b}{a}\right)^{2}}\right]+b\left[\frac{2 \cdot\left(\frac{b}{a}\right)}{1+\left(\frac{b}{a}\right)^{2}}\right]$

$=a\left[\frac{1-\frac{b^{2}}{a^{2}}}{1+\frac{b^{2}}{a^{2}}}\right]+b\left[\frac{\frac{2 b}{a}}{1+\frac{b^{2}}{a^{2}}}\right]$

$=a\left[\dfrac{\frac{a^{2}-b^{2}}{a^{2}}}{\frac{a^{2}+b^{2}}{a^{2}}}\right]+b\left[{\dfrac{\frac{2 b}{a}}{\frac{a^{2}+b^{2}}{a ^2}}}\right]$

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

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

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

$=\frac{a\left(a^{2}+b^{2}\right)}{a^{2}+b^{2}}$

Question 31

Sol :
$1+2 \cos 2 \theta=1+2\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$

$=1+2\left[\frac{1-\left(1+2 \tan ^{2} \phi\right)}{1+\left(1+2 \tan ^{2} \phi\right)}\right]$

$=1+2\left[\frac{1-1-2 \tan ^{2} \phi}{1+1+2 \tan ^{2} \phi}\right]$

$=1-\frac{4 \tan ^{2} \phi}{2+2 \tan ^{2} \phi}$

$=1-\frac{4 \tan ^{2} \phi}{2\left(1+\tan ^{2} \phi\right)}$

$=\frac{1+\tan ^{2} \phi-2 \tan ^{2} \phi}{1+\tan ^{2} \phi}$

$=\frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi}$

$=\cos 2 \phi$

Question 32

Sol :
$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$

$\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=\frac{3\left(\frac{1-\tan ^{2} \beta}{1+\tan ^{2} \beta}\right)-1}{3-\left(\frac{1-\tan ^{2} \beta}{1+\tan ^{2} \beta}\right)}$

$\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=\frac{\frac{3-3 \tan ^{2} \beta-\left(1+\tan ^{2} \beta\right)}{1+\tan ^{2} \beta}}{\frac{3+3+\tan^{2} \theta-\left(1-\tan ^{2} \beta\right)}{1+\tan ^{2} \beta}}$

$\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=\frac{3-3 \tan ^{2} \beta-1-\tan ^{2} \beta}{3+3 \tan ^{2} \beta-1+\tan ^{2} \beta}$

$\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=\frac{2-4 \tan ^{2} \beta}{2+4 \tan ^{2} \beta}$

$\frac{1-\tan ^{2} \alpha}{1+\operatorname{tan}^{2} \alpha}=\frac{2\left(1-2 \tan ^{2} \beta\right)}{2(1+2\tan^{2}\beta)}$

$\left(1-\tan ^{2} \alpha\right)\left(1+2 \tan ^{2} \beta\right)=\left(1+\tan ^{2} \alpha\right)\left(1-2 \tan ^{2} \beta\right)$

$1+2 \tan ^{2} \beta-\tan ^{2} \alpha-2 \tan ^{2} \alpha \tan ^{2} \beta =1-2 \tan ^{2} \beta+\tan ^{2} \alpha-2 \tan ^{2} \alpha \tan^2 \beta$

$-\tan ^{2} \alpha-\tan ^{2} \alpha=-2 \tan ^{2} \beta-2 \tan ^{2} \beta$

$-2 \tan ^{2} \alpha=-4 \tan ^{2} \beta$

$\tan ^{2} \alpha=\frac{4\tan^{2} \beta}{2}$

$\tan \alpha=\sqrt{2 \tan ^{2} \beta}$

$\tan \alpha=\sqrt{2} \tan \beta$

Question 33

Sol :
R.H.S

$\frac{2 \sin 2 \theta}{1+2 \cos 2 \beta}=\frac{\frac{2 \cdot 2 \tan \beta}{1+\tan ^{2} \beta}}{1+2\left(\frac{1-\tan ^{2} \beta}{1+\tan ^{2} \beta}\right)}$

$=\dfrac{\frac{4\tan \beta}{1+\tan ^{2} \beta}}{\frac{1+\tan^{2} \beta+2-2 \tan ^{2} \beta}{1+\tan ^{2} \beta}}$

$=\frac{4 \tan \beta}{3-\tan ^{2} \beta}$


$\because \tan \beta=3 \tan \alpha$

$=\frac{3 \tan \beta+\tan \beta}{3-\tan \beta\tan \beta}$

$=\frac{3 \tan \beta+3 \tan \alpha}{3-3 \tan \alpha \tan \beta}$

$=\frac{3(\tan \alpha+\tan \beta)}{3(1-\tan \alpha \tan \beta)}$

$=\tan (\alpha+\beta)$

Question 34

Sol :

xsinɑ=ycosɑ

$\Rightarrow \frac{\sin\alpha}{\cos \alpha}=\frac{y}{x}$

$\Rightarrow \tan \alpha=\frac{y}{x}$


L.H.S
$\frac{x}{\sec 2 x}+\frac{y}{cosec 2 x}$

$=x \cos 2 \alpha+y \sin 2 \alpha$

$=x\left(\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}\right)+y\left(\frac{2 \tan \alpha}{1+\tan ^{2} \alpha}\right)$

$=x\left[\frac{1-\left(\frac{y}{x}\right)^{2}}{1+\left(\frac{y}{x}\right)^{2}}\right]+y\left[\frac{2 \frac{y}{x}}{1+\left(\frac{y}{2}\right)^{2}}\right]$

$=x\left[\frac{1-\frac{y^{2}}{x^{2}}}{1+\frac{y^{2}}{x^{2}}}\right]+y\left[\frac{\frac{2 y}{2}}{1+\frac{y^{2}}{x^{2}}}\right]$

$=x\left[\frac{\frac{x^{2}-y^{2}}{x^{2}}}{\frac{x^{2}+y^{2}}{x^{2}}}\right]+y\left[\frac{\frac{2 y}{x}}{\frac{x^{2}+y^{2}}{x^{2}}}\right]$

$=x\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)+\frac{2 xy^{2}}{x^{2}+y^{2}}$

$=\frac{x^{3}-x y^{2}+2 x y^{2}}{x^{2}+y^{2}}$

$=\frac{x^{3}+x y^{2}}{x^{2}+y^{2}}$

$=\frac{x\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}=x$

Question 35

Sol :
$\tan \theta=\sec2\alpha$

$\Rightarrow \tan \theta=\frac{1}{\cos 2 \alpha}$

$\tan \theta=\frac{1}{\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}}$

$=\frac{1+\tan ^{2} \alpha}{1-\tan ^{2} \alpha}$


L.H.S

$\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$=\frac{2\left(\frac{1+\tan ^{2} \alpha}{1-\tan ^{2} \alpha}\right)}{1+\left(\frac{1+\tan ^{2} \alpha}{1-\tan ^{2} \alpha}\right)^{2}}$

$\sin 2 \theta=\frac{\frac{2\left(1+\tan ^{2} \alpha\right)}{1-\tan ^{2} \alpha}}{1+\frac{\left(1+\tan ^{2} \alpha\right)^{2}}{\left(1-\tan ^{2} \alpha\right)^{2}}}$

$\sin 2\theta=\dfrac{\frac{2\left(1+\tan ^{2} \alpha\right)}{1-\tan ^{2} \alpha}}{\frac{\left(1-\tan ^{2} \alpha\right)^{2}+\left(1+\tan ^{2} \alpha\right)^{2}}{\left(1-\tan ^{2} \alpha\right)^{2}}}$

$\sin 2 \theta=\frac{2\left(1+\tan ^{2} \alpha\right)\left(1-\tan ^{2} \alpha\right)}{1^{2}-2 \cdot 1 \cdot \tan ^{2} \alpha+\left(\tan ^{2} \alpha\right)^{2}+1^{2}+2 \cdot 1 \cdot \tan ^{2} \alpha +\left(\tan ^{2} \alpha\right)^{2}}$

$\sin 2 \theta=\frac{2\left[1^{2}-\left(\tan ^{2} \alpha\right)^{2}\right]}{2+2 \tan ^{4} \alpha}$

$\sin 2 \theta=\frac{2\left(1-\tan ^{4} \alpha\right)}{2(1+\tan^4 \alpha)}$

$\sin 2 \theta=\frac{1-\tan ^{4} \alpha}{1+\tan ^{4} \alpha}$