Trigonometric Identities Question 30
Question: If $\cos x+\cos y+\cos \alpha=0$ and $\sin x+\sin y+\sin \alpha=0$ then $\cot \left(\frac{x+y}{2}\right)=$
[Karnataka CET 2001]
Options:
A) $ \sin \alpha $
B) $ \cos \alpha $
C) $ \cot \alpha $
D) $\sin \left(\frac{x+y}{2}\right)$
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation $ \cos x+\cos y+\cos \alpha =0 $ and $ \sin x+\sin y+\sin \alpha =0. $
The given equation may be written as $ \cos x+\cos y=-\cos \alpha $ and $ \sin x+\sin y=-\sin \alpha . $
Therefore $ 2\cos ( \frac{x+y}{2} )\cos ( \frac{x-y}{2} )=-\cos \alpha $ ..(i)
$ 2\sin ( \frac{x+y}{2} )\cos ( \frac{x-y}{2} )=-\sin \alpha $ ..(ii)
Divide (i) by (ii), we get $ \frac{2\cos ( \frac{x+y}{2} )\cos ( \frac{x-y}{2} )}{2\sin ( \frac{x+y}{2} )\cos ( \frac{x-y}{2} )} $
$ =\frac{\cos \alpha }{\sin \alpha } $
Þ $ \cot ( \frac{x+y}{2} )=\cot \alpha $ .
BETA
