Differentiation Question 16
Question: If $ y={{\tan }^{-1}}\sqrt{\frac{1+\cos x}{1-\cos x}} $ , then $ \frac{dy}{dx} $ is equal to
[Roorkee 1995]
Options:
A) 0
B) $ -\frac{1}{2} $
C) ½
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ y={{\tan }^{-1}}\sqrt{\frac{1+\cos x}{1-\cos x}}={{\tan }^{-1}}\sqrt{\frac{2{{\cos }^{2}}\frac{x}{2}}{2{{\sin }^{2}}\frac{x}{2}}} $
$ ={{\tan }^{-1}}\cot \frac{x}{2}={{\tan }^{-1}}\tan ( \frac{\pi }{2}-\frac{x}{2} )=\frac{\pi }{2}-\frac{x}{2} $
Therefore $ \frac{dy}{dx}=-\frac{1}{2} $ .
BETA
