Question

If $ f(x)={{\cot }^{-1}}\left[ \sqrt{\frac{1+\sin x}{1-\sin x}} \right],\,0\le x\le \frac{\pi }{4}, $ then $ f'\left( \frac{\pi }{4} \right) $ is equal to

• A. $ \frac{1}{2} $
• B. $ \frac{-1}{2} $
• C. $ \frac{1-{{x}^{2}}}{1+{{x}^{2}}} $
• D. $ \frac{x}{1-{{x}^{2}}} $

Answer 1

Answer: (B)

$ 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} $