Question

$\frac{d}{dx}\sqrt{\left(\frac{1-\sin 2x}{1+\sin 2x}\right)}$is equal to

• A. $se{c}^{2}x$
• B. $-se{c}^2\left(\frac{\pi }{4}–x\right)$
• C. $se{c}^2\left(\frac{\pi }{4}+x\right)$
• D. $se{c}^2\left(\frac{\pi }{4}–x\right)$

Answer 1

The correct option is B

$-se{c}^2\left(\frac{\pi }{4}–x\right)$

Explanation for the correct option:

Step 1. Find the differentiation of the given expression:

Given, $y=\sqrt{\left(\frac{1-\sin 2x}{1+\sin 2x}\right)}$

$\Rightarrow$ $\begin{array}{rcl}y& =& \sqrt{\left(\frac{{\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x}{{\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x}\right)}\end{array}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{sin}\mathit{\theta }\mathbf{}\mathbf{cos}\mathit{\theta }\right]$

$\Rightarrow$ $\begin{array}{rcl}y& =& \sqrt{\left[\frac{{(\cos x–\sin x)}^2}{{(\cos x+\sin x)}^2}\right]}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}y& =& \frac{(\cos x–\sin x)}{(\cos x+\sin x)}\end{array}$

Step 2. Divide numerator and denominator by $\mathbf{cos}\mathbf{}\mathbf{x}$

$y=\frac{(1-\tan x)}{(1+\tan x)}$

$\Rightarrow$$y=\frac{(\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan x)}{(\tan \frac{\mathrm{\pi }}{4}+\tan x)}$ $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\left(\frac{\pi }{4}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$$y=\tan \left(\frac{\mathrm{\pi }}{4}-x\right)$ $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\left(\theta -\varphi \right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{-}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\varphi }}{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\varphi }}\right]$

Differentiate it with respect to $x$, we get

$\begin{array}{rcl}\mathbf{\therefore }\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{-}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathbf{}\left(\frac{\pi }{4}–x\right)\end{array}$

Hence, option (B) is correct.