Question

If $y=co{t}^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$, then $\frac{dy}{dx}$ is equal to ?

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $3$
• D. $1$

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option:

Step 1: Find the value of $y$

Given,

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

On rationalization the function,

$\begin{array}{rcl}y& =& co{t}^{-1}\frac{{\left[(\sqrt{1+\sin x)}+\sqrt{(1–\sin x)}\right]}^2}{\left[(1+\sin x)–(1–\sin x)\right]}\\ & =& co{t}^{-1}\frac{\left[(1+\sin x)+(1–\sin x)+2\sqrt{(1–{\sin }^{2}x)}\right]}{(1+\sin x–1+\sin x)}\\ & =& co{t}^{-1}\frac{(2+2\cos x)}{(2\sin x)}\left[\because {\sin }^{2}A+{\cos }^{2}A=1\right]\\ & =& co{t}^{-1}\frac{2(1+\cos x)}{(2\sin x)}\\ & =& co{t}^{-1}\left(\frac{(1+\cos x)}{\sin x}\right)\\ & =& co{t}^{-1}\left(\frac{2{\cos }^2\left(\frac{x}{2}\right)}{2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)}\right)\left[\because \cos 2A=2{\cos }^{2}A-1,\sin 2A=2\sin A\cos A\right]\\ & =& co{t}^{-1}\left(cot\frac{x}{2}\right)\\ & =& \frac{x}{2}\end{array}$

Step 2: Find the value of $\frac{dy}{dx}$

Now, differentiate with respect to $x$

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

Hence, the correct option is A.