Exercise 6.2
Question 1
(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]
=2 \sin \dfrac{65^{\circ}+25^{\circ}}{2} \cos \dfrac{65^{\circ}-25^{\circ}}{2}
=2sin45°cos20°
=2 \times \dfrac{1}{\sqrt{2}} \cos 20^{\circ}
=√2cos20°
Question 2
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}}
=\frac{\sin 80^{\circ}-\sin 10^{\circ}}{\sin 80^{\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}}
=\frac{\cos 45^{\circ} \sin 35^{\circ}}{\sin 45^{\circ}}
=\frac{\frac{1}{\sqrt{2}} \sin 35^{\circ}}{\frac{1}{\sqrt{2}} \cos 35^{\circ}}
=tan35°
Question 3
साबित करे कि (Prove that) cos80°+cos40°-cos20°=0Sol :
L.H.S
=cos80°+cos40°-cos20°
=2 Cos\frac{ 80^{\circ}+40^{\circ}}{2} \cos \frac{80^{\circ}-40^{\circ}}{2}-\cos 20^{\circ}
=2cos60°.cos20°-cos20°
=2 \times \dfrac{1}{2} \cos 20^{\circ}-\cos 20^{\circ}
=0
Question 4
साबित करे कि (Prove that) sin10°+sin20°+sin40°+sin50°=sin70°+sin80°Sol :
L.H.S
=sin10°+sin20°+sin40°+sin50°
=2sin30°cos20°+2sin30°cos10°
=(sin90°-20°)+(sin90°-10°)
=sin70°+sin80°
Question 5
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
Sol :
L.H.S
=\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}
Question 6
साबित करे कि (Prove that)
(i) \frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x
Sol :
L.H.S
∵\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
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
=\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)
[]
\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}
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
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]
\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}
\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}
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
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
\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}
\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}
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
y \sin \phi=x \sin (2 \theta+\phi)
\frac{y}{x}=\frac{\sin (2 \theta+\phi)}{\sin \phi}
[]
\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
\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)}
[]
\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
\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)}
[]
\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
(i)
\tan (\theta+\phi)=3 \tan \theta
\frac{\tan (\theta+\phi)}{\tan \theta}=\frac{3}{1}
[]
\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
[]
[∵ 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
∵\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]
=2sin4x.2cos2x
=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
Question 10
(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)
[]
\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
[]\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
[]\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 \thetaSol :
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 \thetaSol :
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+\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}
[]
\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)}
[]
\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)}
[]
\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}
[]
\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
[]
[∵ 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
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