Trigonometric Identities Question 193
Question: If A, B, C are angles of a triangle, then 2sin $ \frac{A}{2}\cos ec,\frac{B}{2}\sin \frac{C}{2}-\sin A\cot \frac{B}{2}-\cos A $ is
Options:
A) independent of A, B,
B) function of A, B
C) function of C
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ = $ $ 2\sin \frac{A}{2}\cos ec\frac{B}{2}( \sin \frac{C}{2}-\cos \frac{A}{2}\cos \frac{B}{2} )-\cos A $ $ =2\sin \frac{A}{2}\cos ec\frac{B}{2}\times ( \cos \frac{A+B}{2}-Cos\frac{A}{2}\cos \frac{B}{2} )-\cos A $ $ = $ $ 2\sin \frac{A}{2}\cos ec\frac{B}{2}\times ( -\sin \frac{A}{2}\sin \frac{B}{2} )-\cos A $ $ =-2{{\sin }^{2}}\frac{A}{2}-\cos A=-1 $
BETA
