Differentiation Question 104
Question: $ \frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)= $
[AISSE 1985, 87; DSSE 1982, 84]
Options:
A) 1
B) 1/2
C) $ \cos x $
D) $ \sec x $
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Answer:
Correct Answer: B
Solution:
$ \frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)=\frac{d}{dx}{{\tan }^{-1}}( \frac{1+\sin x}{\cos x} ) $
$ =\frac{d}{dx}{{\tan }^{-1}}( \frac{\sin ( \frac{x}{2} )+\cos ( \frac{x}{2} )}{\cos ( \frac{x}{2} )-\sin ( \frac{x}{2} )} )=\frac{d}{dx}( \frac{\pi }{4}+\frac{x}{2} )=\frac{1}{2} $ .
BETA
