Question

If $ \sin \theta =3\sin (\theta +2\alpha ), $ then the value of $ \tan (\theta +\alpha )+2\tan \alpha $ is

• A. $ 3 $
• B. $ 2 $
• C. $ -1 $
• D. $ 0 $
• E. $ 1 $

Answer 1

Answer: (D)

Given, $ \sin \theta =3\sin (\theta +2\alpha ) $
$ \Rightarrow $ $ \frac{\sin \theta }{\sin (\theta +2\alpha )}=\frac{3}{1} $ Use componendo and dividendo formula,
$=\frac{\sin \theta +\sin (\theta +2\alpha )}{\sin \theta -\sin (\theta +2\alpha )}=\frac{3+1}{3-1} $
$=-\frac{2\sin (\theta +\alpha ).\cos \alpha }{2\cos (\theta +\alpha ).\sin \alpha }=2 $
$ \Rightarrow $ $ -\tan (\theta +\alpha ).\cot \alpha =2 $
$ \Rightarrow $ $ \tan (\theta +\alpha )=-2\tan \alpha $
$ \Rightarrow $ $ \tan (\theta +\alpha )=2\tan \alpha =0 $