Trigonometric Identities Question 56
Question: If $ \cos A=m\cos B, $ then
[MNR 1990]
Options:
A) $ \cot \frac{A+B}{2}=\frac{m+1}{m-1}\tan \frac{B-A}{2} $
B) $ \tan \frac{A+B}{2}=\frac{m+1}{m-1}\cot \frac{B-A}{2} $
C) $ \cot \frac{A+B}{2}=\frac{m+1}{m-1}\tan \frac{A-B}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ \cos A=m\cos B,\Rightarrow \frac{m}{1}=\frac{\cos A}{\cos B} $
$ \Rightarrow \frac{m+1}{m-1}=\frac{\cos A+\cos B}{\cos A-\cos B}=\frac{2\cos ( \frac{A+B}{2} )\cos ( \frac{B-A}{2} )}{2\sin ( \frac{A+B}{2} )\sin ( \frac{B-A}{2} )} $ $ =\cot ,( \frac{A+B}{2} ),\cot ,( \frac{B-A}{2} ) $ Hence, $ \cot ,( \frac{A+B}{2} )=\frac{m+1}{m-1}\tan \frac{B-A}{2} $ .
BETA
