Question

If $m\sin \theta =n\sin (\theta +2\alpha )$, then $\tan (\theta +\alpha ).\cot \alpha$ equal to

• A. $\frac{1-n}{1+n}$
• B. $\frac{m+n}{m-n}$
• C. $\frac{m-n}{m+n}$
• D. $\frac{1+n}{1-n}$

Answer 1

The correct option is B $\frac{m+n}{m-n}$

Given,

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

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

Apply componendo and dividendo

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

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

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