Applications Of Derivatives Question 187
Question: $ \frac{d}{dx}{{\tan }^{-1}}[ \frac{\cos x-\sin x}{\cos x+\sin x} ]= $
[AISSE 1985, 87; DSSE 1982,84; MNR 1985; Karnataka CET 2002; RPET 2002, 03]
Options:
A) $ \frac{1}{2(1+x^{2})} $
B) $ \frac{1}{1+x^{2}} $
C) 1
D) - 1
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{d}{dx}{{\tan }^{-1}}[ \frac{\cos x-\sin x}{\cos x+\sin x} ] $
$ =\frac{d}{dx}{{\tan }^{-1}}[ \tan ( \frac{\pi }{4}-x ) ]=-1 $ .
BETA
