Question

If $m\sin \theta =n\sin (\theta +2\alpha )$, then
prove that $\tan (\theta +\alpha )\cot \alpha =\frac{m+n}{m-n}$

Answer 1

Given:
$m\sin \theta =n\sin (\theta +2\alpha )$
$\Rightarrow \frac{\sin (\theta +2\alpha )}{\sin \theta }=\frac{m}{n}$
$\Rightarrow \frac{\sin (\theta +2\alpha )+\sin \theta }{\sin (\theta +2\alpha )-\sin \theta }=\frac{m+n}{m-n}$
$[\because \frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}]$
$\Rightarrow \frac{2\sin (\theta +\alpha )\cos \alpha }{2\cos (\theta +\alpha )\sin \alpha }=\frac{m+n}{m-n}$
$\left[{}_{and\sin C–\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}}^{\because \sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}}\right]$
$\Rightarrow \frac{2\sin (\theta +\alpha )\cos \alpha }{2\cos (\theta +\alpha )\sin \alpha }=\frac{m+n}{m-n}$
$\Rightarrow \tan (\theta +\alpha )\cot \alpha =\frac{m+n}{m-n}$
$[\therefore \cot \theta =\frac{\cos \theta }{\sin \theta }]$

Hence, Proved.