Question

If $y=\frac{\tan x+\cot x}{\tan x-\cot x}$, then $\frac{dy}{dx}=$

• A. $2\tan 2x\sec 2x$
• B. $\tan 2x\sec 2x$
• C. $-\tan 2x\sec 2x$
• D. $-2\tan 2x\sec 2x$

Answer 1

The correct option is D

$-2\tan 2x\sec 2x$

Explanation for the correct option.

Step 1. Simplify the given equation.

Using the relation $\cot x=\frac{1}{\tan x}$, simplify the equation $y=\frac{\tan x+\cot x}{\tan x-\cot x}$.

$\begin{array}{rcl}y& =& \frac{\tan x+{\displaystyle \frac{1}{\tan x}}}{\tan x-{\displaystyle \frac{1}{\tan x}}}\\ & =& \frac{{\tan }^{2}x+1}{{\tan }^{2}x-1}\\ & =& \frac{1+{\tan }^{2}x}{-\left(1-{\tan }^{2}x\right)}\\ & =& -\frac{1}{\cos 2x}\left[\because \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\right]\\ & =& -\sec 2x\end{array}$

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

Differentiate both sides of the equation $y=-\sec 2x$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}(-\sec 2x)\\ & =& -\sec 2x\tan 2x\times 2\left[\because \frac{d}{dx}\left(secx\right)=secx\tan x\right]\\ & =& -2\sec 2x\tan 2x\end{array}$

Hence, the correct option is D.