Browse Questions

# If $\sin\theta=n\sin(\theta+2\alpha)$ then $tan(\theta+\alpha)$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{1+n}{2-n}\normalsize\tan\alpha&(b)\large\frac{1-n}{1+n}\normalsize\tan\alpha\\(c)\;\tan\alpha&(d)\;\large\frac{1+n}{1-n}\normalsize\tan\alpha\end{array}$

Can you answer this question?

We have
$\sin\theta=n\sin(\theta+2\alpha)$
$\Rightarrow \large\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{1}{n}$
$\Rightarrow \large\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}$
$\Rightarrow \large\frac{1+n}{1-n}$
$\Rightarrow \large\frac{2\sin(\theta+\alpha).\cos\alpha}{2\sin\alpha\cos(\theta+\alpha)}=\frac{1+n}{1-n}$