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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

tan⁡2xtan⁡x\tan 2x\tan x

B

tan⁡3xtan⁡x\tan 3x\tan x

C

sec⁡2x\sec^{2}x

D

sec⁡xtan⁡x\sec x\tan x

Answer

sec⁡2x\sec^{2}x

Explanation

Solution

Let y=tan⁡22x−tan⁡2x1−tan⁡22xtan⁡2xy = \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(2x - x)\tan(2x + x) = tan⁡xtan⁡3x\tan x\tan 3x.

∴ ddx[y.cot⁡3x]=ddx[tan⁡x]=sec⁡2x\frac{d}{dx}\lbrack y.\cot 3x\rbrack = \frac{d}{dx}\lbrack\tan x\rbrack = \sec^{2}x.