Differentiation Question 151
Question: $ \frac{d}{dx}( {{\tan }^{-1}}\frac{\cos x}{1+\sin x} )= $
[AISSE 1984, 85; MNR 1983; RPET 1997]
Options:
A) $ -\frac{1}{2} $
B) $ \frac{1}{2} $
C) $ -1 $
D) 1
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Answer:
Correct Answer: A
Solution:
$ \frac{d}{dx}[ {{\tan }^{-1}}( \frac{\cos x}{1+\sin x} ) ] $
$ =\frac{d}{dx}[ {{\tan }^{-1}}( \frac{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} ) ] $
$ =\frac{d}{dx}[ {{\tan }^{-1}}( \frac{1-\tan ( \frac{x}{2} )}{1+\tan ( \frac{x}{2} )} ) ]=\frac{d}{dx}[ {{\tan }^{-1}}\tan ( \frac{\pi }{4}-\frac{x}{2} ) ]=-\frac{1}{2} $
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