A) \[\frac{1}{2}\]
B) \[2\]
C) \[-2\]
D) \[-\frac{1}{2}\]
E) \[-1\]
Correct Answer: D
Solution :
Given that, \[y={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right)\] \[={{\tan }^{-1}}\left( \frac{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \right)\] \[={{\tan }^{-1}}\left( \frac{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}} \right)\] \[={{\tan }^{-1}}\left( \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} \right)\] \[={{\tan }^{-1}}\left( \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \right)\] \[={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}-\frac{x}{2} \right) \right)\] \[=\frac{\pi }{4}-\frac{x}{2}\] \[\therefore \] \[y=\frac{\pi }{4}-\frac{x}{2}\] Differentiating it w.r.t.,\[x\]we get \[\frac{dy}{dx}=-\frac{1}{2}\]
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