Question

$\frac{d}{dx}\left(\frac{secx+\tan x}{secx–\tan x}\right)=?$

• A. $\frac{2\cos x}{{(1-\sin x)}^2}$
• B. $\frac{\cos x}{{(1-\sin x)}^2}$
• C. $\frac{2\cos x}{(1-\sin x)}$
• D. None of these

Answer 1

The correct option is A

$\frac{2\cos x}{{(1-\sin x)}^2}$

Find the differentiation of the given function:

Given, $y=\left(\frac{secx+\tan x}{secx–\tan x}\right)$

$=\left(\frac{1+\sin x}{1-\sin x}\right)$

Differentiate it with respect to $x$, we get

$\begin{array}{rcl}\therefore \frac{dy}{dx}& =& \frac{\left[(1-\sin x)(\cos x)+(1+\sin x)(\cos x)\right]}{{(1-\sin x)}^2}\\ & =& \frac{\left[\cos x–(\sin x\cos x)+\cos x+(\sin x\cos x)\right]}{{(1-\sin x)}^2}\\ & =& \frac{\mathbf{2}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{x}}{{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{)}}^{\mathbf{2}}}\end{array}$ ; $\left[\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(\frac{u}{v}\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{v}{\displaystyle \frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{u}{\displaystyle \frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}}}{{\mathbf{v}}^{\mathbf{2}}}\right]$

Hence, option (A) is correct.