Question

If $\sin \theta =n\sin (\theta +2\alpha )$ then $\tan (\theta +\alpha )$ =

• A. $\frac{1+n}{1-n}\tan \alpha$
• B. $\frac{1-n}{1+n}\tan \alpha$
• C. $\tan \alpha$
• D. None

Answer 1

The correct option is A $\frac{1+n}{1-n}\tan \alpha$

$\sin \theta =n\sin (\theta +2\alpha )$

$\frac{\sin \theta }{\sin (\theta +2\alpha )}=n$

apply componendo dividendo rule,

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

apply $\sin C+\sin D$ and $\sin C-\sin D$

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

$\frac{\tan (\theta +\alpha )}{\tan \alpha }=\frac{n+1}{n-1}$

$\tan (\theta +\alpha )=\frac{n+1}{n-1}\tan \alpha$