Solveeit Logo
Login

Question

Question: \[\frac{d}{dx}\left\lbrack \left( \frac{\tan^{2}2x - \tan^{2}x}{1 - \tan^{2}2x\tan^{2}x} \right)\cot...

ddx[(tan⁡22x−tan⁡2x1−tan⁡22xtan⁡2x)cot⁡3x]\frac{d}{dx}\left\lbrack \left( \frac{\tan^{2}2x - \tan^{2}x}{1 - \tan^{2}2x\tan^{2}x} \right)\cot 3x \right\rbrack

A

tan2x tanx

B

tan3x tanx

C

sec2x

D

secx tanx

Explanation

Solution

Let y = tan⁡22x−tan⁡2x1−tan⁡22xtan⁡2x\frac{\tan^{2}2x - \tan^{2}x}{1 - \tan^{2}2x\tan^{2}x}

= (tan⁡2x−tan⁡x)(1+tan⁡2xtan⁡x)(tan⁡2x+tan⁡x)(1−tan⁡2xtan⁡x)\frac{(\tan 2x - \tan x)}{(1 + \tan 2x\tan x)}\frac{(\tan 2x + \tan x)}{(1 - \tan 2x\tan x)}

= tan(2x – x) tan(2x + x) = tan x tan 3x

\ ddx[y.cot⁡3x]\frac{d}{dx}\lbrack y.\cot 3x\rbrack = ddx[tan⁡x]\frac{d}{dx}\lbrack\tan x\rbrack = sec2x.