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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 40 of 61
Marks: +1, -0
tan 4x = 4tan⁡x(1−tan⁡2x)1−6tan⁡2x+tan⁡4x\frac{4\tan x(1-\tan^2 x)}{1-6\tan^2 x + \tan^4 x}
Solution:  
We have, L.H.S. = tan 4x = tan2(2x)
= 2tan⁡2x1−tan⁡2x\frac{2\tan 2x}{1-\tan^2 x}
Since tan 2θ = 2tan⁡θ1−tan⁡2θ\frac{2\tan\theta}{1-\tan^2\theta}
= 2(2tan⁡x1−tan⁡2x)1−(2tan⁡x1−tan⁡2x)2\frac{2\left(\frac{2\tan x}{1-\tan^2 x}\right)}{1-\left(\frac{2\tan x}{1-\tan^2 x}\right)^2} = 4tan⁡x1−tan⁡2x\frac{4\tan x}{1-\tan^2 x} × (1−tan⁡2x)2(1−tan⁡2x)2−4tan⁡2x\frac{(1-\tan^2 x)^2}{(1-\tan^2 x)^2-4\tan^2 x}
= 4tan⁡x(1−tan⁡2x)1+tan⁡4x−2tan⁡2x−4tan⁡2x\frac{4\tan x(1-\tan^2 x)}{1+\tan^4 x-2\tan^2 x-4\tan^2 x} = 4tan⁡(1−tan⁡2x)1−6tan⁡2x+tan⁡4x\frac{4\tan(1-\tan^2 x)}{1-6\tan^2 x+\tan^4 x} = R.H.S.
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