Question

$y={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$ then $\frac{dy}{dx}=$

• A. $1/2$
• B. $2/3$
• C. $3$
• D. $1$

Answer 1

The correct option is A $1/2$

$y=co{t}^{-1}\left[\frac{\sqrt{1-sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$

Rationalising,

$=co{t}^{-1}\left[\frac{(\sqrt{1-sinx}+\sqrt{1-sinx}{)}^2}{\sqrt{(1+sinx{)}^2}-\sqrt{(1-sinx{)}^2}}\right]$

$=co{t}^{-1}\left[\frac{(\sqrt{1+sinx}+\sqrt{1-sin}{)}^2}{1+sinx-1+sinx}\right]$

$=co{t}^{-1}\left[\frac{(1+sinx+1-sinx+2\sqrt{1-sinx})}{2sinx}\right]$

$=co{t}^{-1}\left[\frac{2+2cos}{sinx}\right]$ $=[\because \frac{1+cosx}{sinx}=\frac{sinx}{1-cosx}]$

$=co{t}^{-1}\left(\frac{sinx}{1-cosx}\right)$

$=co{t}^{-1}\left[\frac{2sinx/2cosx/2}{1-(1-2sin\frac{2x}{2})}\right]=co{t}^1\left[\frac{2/sinx/2cosx/2}{2si{n}^{2}x/2}\right]$

$y\Rightarrow co{t}^{-1}\left[cot\right(\frac{x}{2}\left)\right]=\frac{x}{2}$

$\because y=\frac{x}{2}$

$\frac{dy}{dx}=\frac{1}{2}$

Option A is correct