Trigonometric Equations Question 35
Question: In $ \Delta ABC $ , $ (b-c)\cot \frac{A}{2}+(c-a)\cot \frac{B}{2}+(a-b) $ $ \cot \frac{C}{2} $ is equal to
[WB JEE 1989]
Options:
A) 0
B) 1
C) $ \pm 1 $
D) 2
Show Answer
Answer:
Correct Answer: A
Solution:
- $ (b-c)\cot \frac{A}{2}=k(\sin B-\sin C)\cot \frac{A}{2} $ $ =2k\cos \frac{B+C}{2}\sin \frac{B-C}{2}\cot \frac{A}{2} $ $ =2k\sin \frac{A}{2},.\sin \frac{B-C}{2}\text{. }\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} $ $ =2k\sin ( \frac{B-C}{2} )\sin ( \frac{B+C}{2} )=2k( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} ) $ or we get L.H.S. = $ \Sigma 2k( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} )=0 $ .
BETA
