A) \[\tan \,x=\tan \alpha \]
B) \[\tan \,x={{\tan }^{2}}\alpha \]
C) \[\tan x={{\tan }^{3}}\alpha \]
D) \[\tan x=3\tan \alpha \]
Correct Answer: C
Solution :
Given \[\frac{\sin (x+3\alpha )}{\sin (\alpha -x)}=3\] Applying componendo and dividendo, we get \[\Rightarrow \]\[\frac{\sin (x+3\alpha )+\sin (\alpha -x)}{\sin (x+3\alpha )-\sin (\alpha -x)}=\frac{3+1}{3-1}\] \[\Rightarrow \]\[\frac{2\sin 2\alpha \cos (\alpha +x)}{2\cos 2\alpha \sin (\alpha +x)}=2\] \[\Rightarrow \]\[\frac{\tan 2\alpha }{\tan (\alpha +x)}=2\] \[\Rightarrow \]\[\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\times \frac{(1-\tan \alpha \tan x)}{(\tan \alpha +\tan x)}=2\] \[\Rightarrow \]\[\tan \alpha -{{\tan }^{2}}\alpha \tan x=\tan \alpha +\tan x\] \[-{{\tan }^{3}}\alpha -{{\tan }^{2}}\alpha \tan x\] \[\Rightarrow \]\[\tan x={{\tan }^{3}}\alpha \]
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