1. Tardigrade
  2. Question
  3. Mathematics
  4. If sin A=n sin B , then (n-1/n+1) tan (A+B/2) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $ sin\,A=n\,sin\,B $ , then $ \frac{n-1}{n+1}\,tan\, \frac{A+B}{2} $ is equal to

UPSEEUPSEE 2007

Solution:

Given, $\sin \,A=n \sin \,B$
$\Rightarrow \, \frac{n}{1}=\frac{\sin A}{\sin B}$
Applying componendo and dividendo, we get
$\frac{n-1}{n+1} =\frac{\sin A-\sin B}{\sin A+\sin B} $
$=\frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)} $
$\Rightarrow \frac{n-1}{n+1}= \tan \left(\frac{A-B}{2}\right) \cot \left(\frac{A+B}{2}\right)$
$\Rightarrow \,\frac{n-1}{n+1} \tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{A-B}{2}\right)$