Trigonometric Identities Question 130
Question: If $ \sin A=n\sin B, $ then $ \frac{n-1}{n+1}\tan ,\frac{A+B}{2}= $
Options:
A) $ \sin \frac{A-B}{2} $
B) $ \tan \frac{A-B}{2} $
C) $ \cot \frac{A-B}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \sin A=n\sin B\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B} $
$ \Rightarrow \frac{n-1}{n+1}=\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}} $ $ =\tan \frac{A-B}{2}\cot \frac{A+B}{2} $
$ \Rightarrow \frac{n-1}{n+1}\tan ( \frac{A+B}{2} )=\tan \frac{A-B}{2} $ .
BETA
