Question

If $y=\left(\frac{{\tan }^{2}2x-{\tan }^{2}x}{1-{\tan }^{2}2x{\tan }^{2}x}\right)\cot 3x,$ then $y^{\prime }\left(\frac{\pi }{4}\right)=$

Answer 1

We have,
$y=\left(\frac{{\tan }^{2}2x-{\tan }^{2}x}{1-{\tan }^{2}2x{\tan }^{2}x}\right)\cot 3x$

$=[\frac{(\tan 2x-\tan x)}{(1+\tan 2x\tan x)}\times \frac{(\tan 2x+\tan x)}{(1-\tan 2x\tan x)}]\cot 3x$
$=\tan (2x-x)\tan (2x+x)\cot 3x$
$=\tan x\tan 3x\cot 3x$
$=\tan x$
$\Rightarrow y^{\prime }=\frac{d}{dx}\left[\tan x\right]={\sec }^{2}x$

$\therefore y^{\prime }\left(\frac{\pi }{4}\right)={\sec }^2\left(\frac{\pi }{4}\right)=2$