Question

If $y=\left(\frac{1+\tan x}{1-\tan x}\right)$, then $\frac{dy}{dx}=$

• A. $\frac{se{c}^{2}x}{{\left(1-\tan x\right)}^2}$
• B. $\frac{2se{c}^{2}x}{{\left(1-\tan x\right)}^2}$
• C. $\frac{2se{c}^{2}x}{{\left(1+\tan x\right)}^2}$
• D. None of these

Answer 1

The correct option is B

$\frac{2se{c}^{2}x}{{\left(1-\tan x\right)}^2}$

Explanation for the correct answer.

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

$\begin{array}{rcl}& \Rightarrow & \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1+\tan x}{1-\tan x}\right)\\ & =& \frac{\left(1-\tan x\right)\left(0+se{c}^{2}x\right)-\left(1+\tan x\right)\left(0-se{c}^{2}x\right)}{{\left(1-\tan x\right)}^2}\left[Usingquotientrule\right]\\ & =& \frac{\left(1-\tan x\right)\left(se{c}^{2}x\right)+\left(1+\tan x\right)\left(se{c}^{2}x\right)}{{\left(1-\tan x\right)}^2}\\ & =& \frac{\left(1-\tan x+1+\tan x\right)\left(se{c}^{2}x\right)}{{\left(1-\tan x\right)}^2}\\ & =& \frac{2se{c}^{2}x}{{\left(1-\tan x\right)}^2}\end{array}$

Hence, option B is correct