Question

If $\sin (\theta +\varphi )=n\sin (\theta -\varphi ),n\ne 1$, then the value of $\frac{\tan \theta }{\tan \varphi }$ is equal to

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

Answer 1

The correct option is B $\frac{n+1}{n-1}$

$\sin (\theta +\varphi )=n\sin (\theta -\varphi )$
$\Rightarrow \frac{\sin (\theta +\varphi )}{\sin (\theta -\varphi )}=n$
Applying componendo and dividendo, we get
$\frac{\sin (\theta +\varphi )+\sin (\theta -\varphi )}{\sin (\theta +\varphi )-\sin (\theta -\varphi )}=\frac{n+1}{n-1}$
$\Rightarrow \frac{\left[2\sin \left(\frac{\theta +\varphi +\theta -\varphi }{2}\right)\cos \left(\frac{\theta +\varphi -\theta +\varphi }{2}\right)\right]}{\left[2\cos \left(\frac{\theta +\varphi +\theta -\varphi }{2}\right)\sin \left(\frac{\theta +\varphi -\theta +\varphi }{2}\right)\right]}=\frac{n+1}{n-1}$
$\Rightarrow \frac{\sin \theta \cos \varphi }{\cos \theta \sin \varphi }=\frac{n+1}{n-1}$
$\Rightarrow \frac{\tan \theta }{\tan \varphi }=\frac{n+1}{n-1}$