Question

If $m\sin \theta =n\sin \left(\theta +2\alpha \right)$, prove that $\tan \left(\theta +\alpha \right)\cot \alpha =\frac{m+n}{m-n}$. [NCERT EXEMPLAR]

Answer 1

Given: $m\sin \theta =n\sin \left(\theta +2\alpha \right)$

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

Applying componendo and dividendo, we get

$$\begin{gathered} \frac{m+n}{m-n}=\frac{\sin \left(\theta +2\alpha \right)+\sin \theta }{\sin \left(\theta +2\alpha \right)-\sin \theta } \\ \Rightarrow \frac{m+n}{m-n}=\frac{2\sin \left({\displaystyle \frac{\theta +2\alpha +\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta +2\alpha -\theta }{2}}\right)}{2\sin \left({\displaystyle \frac{\theta +2\alpha -\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta +2\alpha +\theta }{2}}\right)} \\ \Rightarrow \frac{m+n}{m-n}=\frac{\sin \left(\theta +\alpha \right)\cos \alpha }{\sin \alpha \cos \left(\theta +\alpha \right)} \\ \Rightarrow \frac{m+n}{m-n}=\tan \left(\theta +\alpha \right)\cot \alpha \end{gathered}$$

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