A) 2
B) -1
C) 0
D) -2
Correct Answer: B
Solution :
Let \[u={{\tan }^{-1}}\left( \frac{\sin x}{1+\cos x} \right),v={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right)\] \[u={{\tan }^{-1}}\left( \frac{2\sin x/2.\cos x/2}{1+2{{\cos }^{2}}x/2-1} \right)\] \[={{\tan }^{-1}}(\tan x/2)=x/2\] \[v={{\tan }^{-1}}\left( \frac{\cos x}{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \right)\] \[={{\tan }^{-1}}\left( \frac{\left( {{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2} \right)}{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}} \right)\] \[={{\tan }^{-1}}\left( \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} \right)={{\tan }^{-1}}\left( \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \right)\] \[=\left[ {{\tan }^{-1}}({{\tan }^{-1}}\left( \frac{\pi }{4}-\frac{\pi }{2} \right) \right]=\left( \frac{\pi }{4}-\frac{x}{2} \right)\] Now, \[\frac{du}{dx}=\frac{1}{2}\] and \[\frac{dv}{dx}=\frac{1}{2}\] \[\Rightarrow \] \[\frac{du}{dv}=\frac{du}{dx}\times \frac{dx}{dv}=\left( \frac{1}{2} \right)\times (-2)=-1\]
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