A) \[\frac{1}{2}\]
B) \[\frac{-1}{2}\]
C) \[\frac{1-{{x}^{2}}}{1+{{x}^{2}}}\]
D) \[\frac{x}{1-{{x}^{2}}}\]
Correct Answer: B
Solution :
\[f(x)={{\cot }^{-1}}\left[ \frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}} \right]\] \[\Rightarrow \] \[f'(x)=\frac{-1}{1+\left( \frac{1+\sin x}{1-\sin x} \right)}\,.\,\,\frac{1}{2\sqrt{\frac{1+\sin x}{1-\sin x}}}\] \[\frac{[(1-\sin x)\,\cos \,x-(1+\sin x)\,(-\cos \,x)}{{{(1-\sin \,x)}^{2}}}\] \[=\frac{-(1-\sin x)\sqrt{1-\sin x}}{2.2\,\sqrt{1+\sin x}}\,\,\left[ \frac{2\,\cos x}{{{(1-\sin x)}^{2}}} \right]\] \[=-\frac{1}{2}\,\sqrt{\frac{1-\sin x}{1+\sin x}}.\frac{\cos \,x}{(1-\sin x)}\] \[f'(x)=-\frac{\cos \,x}{2\sqrt{(1+\sin x)\,(1-\,\sin x)}}\] \[=-\frac{\cos \,x}{2.\cos \,x}=-\frac{1}{2}\] \[\Rightarrow \] \[f'\left( \frac{\pi }{4} \right)=-\frac{1}{2}\] Alternate \[f(x)={{\cot }^{-1}}\,\left[ \sqrt{\frac{1+\sin \,\,x}{1-sin\,x}} \right]\] \[f(x)={{\cot }^{-1}}\,\left[ \sqrt{\frac{{{\left( \cos \,\frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}}{{{\left( \cos \,\frac{x}{2}+sn\,\frac{x}{2} \right)}^{2}}}} \right]\] \[f(x)={{\cot }^{-1}}\left( \frac{\cos \,\frac{x}{2}+\sin \frac{x}{2}}{\cos \,\frac{x}{2}-\sin \frac{x}{2}} \right)\] \[f(x)={{\cot }^{-1}}\left[ \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \right]\] \[={{\cot }^{-1}}\left[ \tan \left( \frac{\pi }{4}+\frac{x}{2} \right) \right]\] \[={{\cot }^{-1}}\left[ \cot \,\left\{ \frac{\pi }{2}-\left( \frac{\pi }{4}+\frac{x}{2} \right) \right\} \right]=\frac{\pi }{4}-\frac{x}{2}\] \[f'(x)=-\frac{1}{2}\] \[\Rightarrow \] \[f'\left( \frac{\pi }{4} \right)=-\frac{1}{2}\]
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