Question

$\frac{d}{dx}\left[\left(\frac{{\tan }^2\left(2x\right)-{\tan }^2\left(x\right)}{1-{\tan }^2\left(2x\right){\tan }^2\left(x\right)}\right)\cot \left(3x\right)\right]=$

• A. ${\sec }^2\left(x\right)$
• B. $\sec \left(x\right)\tan \left(x\right)$
• C. $\tan \left(2x\right)\tan \left(x\right)$
• D. none of these

Answer 1

The correct option is A

${\sec }^2\left(x\right)$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{d}{dx}\left[\left(\frac{{\tan }^2\left(2x\right)-{\tan }^2\left(x\right)}{1-{\tan }^2\left(2x\right){\tan }^2\left(x\right)}\right)\cot \left(3x\right)\right]$

$$\begin{gathered} =\frac{d}{dx}\left[\left(\frac{\left(\tan \left(2x\right)-\tan \left(x\right)\right)\left(\tan \left(2x\right)+\tan \left(x\right)\right)}{\left(1-\tan \left(2x\right)\tan \left(x\right)\right)\left(1+\tan \left(2x\right)\tan \left(x\right)\right)}\right)\cot \left(3x\right)\right] \\ =\frac{d}{dx}\left[\tan \left(2x-x\right)\tan \left(2x+x\right)\cot \left(3x\right)\right] \\ =\frac{d}{dx}\left[\tan \left(x\right)\tan \left(3x\right)\cot \left(3x\right)\right] \\ =\frac{d}{dx}\left[\tan \left(x\right)\tan \left(3x\right)\frac{1}{\tan \left(3x\right)}\right] \\ =\frac{d}{dx}\left[\tan \left(x\right)\right] \end{gathered}$$

Step-2: Differentiate w.r.t. $x$

$$\begin{gathered} =\frac{d}{dx}\left(\tan \left(x\right)\right) \\ ={\sec }^2\left(x\right) \end{gathered}$$

Hence, the correct answer is option $\left(A\right)$