Question

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

$f(x)={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}\right]$
$\Rightarrow$ $f^{\prime }(x)=\frac{-1}{1+\left(\frac{1+\sin x}{1-\sin x}\right)}.\frac{1}{2\sqrt{\frac{1+\sin x}{1-\sin x}}}$
$\frac{[(1-\sin x)\cos x-(1+\sin x)(-\cos x)}{{(1-\sin x)}^2}$
$=\frac{-(1-\sin x)\sqrt{1-\sin x}}{2.2\sqrt{1+\sin x}}\left[\frac{2\cos x}{{(1-\sin x)}^2}\right]$
$=-\frac{1}{2}\sqrt{\frac{1-\sin x}{1+\sin x}}.\frac{\cos x}{(1-\sin x)}$
$f^{\prime }(x)=-\frac{\cos x}{2\sqrt{(1+\sin x)(1-\sin x)}}$
$=-\frac{\cos x}{2.\cos x}=-\frac{1}{2}$
$\Rightarrow$ $f^{\prime }\left(\frac{\pi }{4}\right)=-\frac{1}{2}$
Alternate $f(x)={\cot }^{-1}\left[\sqrt{\frac{1+\sin x}{1-sinx}}\right]$
$f(x)={\cot }^{-1}\left[\sqrt{\frac{{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}^2}{{\left(\cos \frac{x}{2}+sn\frac{x}{2}\right)}^2}}\right]$
$f(x)={\cot }^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)$
$f(x)={\cot }^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]$
$={\cot }^{-1}\left[\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right]$
$={\cot }^{-1}\left[\cot \left\{\frac{\pi }{2}-\left(\frac{\pi }{4}+\frac{x}{2}\right)\right\}\right]=\frac{\pi }{4}-\frac{x}{2}$
$f^{\prime }(x)=-\frac{1}{2}$
$\Rightarrow$ $f^{\prime }\left(\frac{\pi }{4}\right)=-\frac{1}{2}$