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Exercise 3.3 · Q23

Q.Prove that tan⁡4x=4tan⁡x (1−tan⁡2x)1−6tan⁡2x+tan⁡4x\tan 4x = \dfrac{4\tan x\,(1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}.

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We prove the identity by repeatedly applying the double-angle formula for tangent: first to get tan⁡2x\tan 2x, then again to express tan⁡4x\tan 4x in terms of tan⁡x\tan x, and finally simplifying to the given form.

Why this approach works.

The identity involves tan⁡4x\tan 4x expressed purely in terms of tan⁡x\tan x. The natural tool is the double-angle formula for tangent:

tan⁡2θ=2tan⁡θ1−tan⁡2θ.\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}.

If we set θ=2x\theta = 2x, then tan⁡4x=tan⁡(2⋅2x)=2tan⁡2x1−tan⁡22x\tan 4x = \tan(2 \cdot 2x) = \frac{2\tan 2x}{1 - \tan^2 2x}.

Now substitute tan⁡2x\tan 2x in terms of tan⁡x\tan x, and simplify the resulting rational expression. The algebra is straightforward but needs careful handling of squares and signs.


Step-by-step proof

  1. Write tan⁡4x\tan 4x using the double-angle formula. Let θ=2x\theta = 2x. Then

tan⁡4x=tan⁡(2θ)=2tan⁡θ1−tan⁡2θ=2tan⁡2x1−tan⁡22x.\tan 4x = \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2\tan 2x}{1 - \tan^2 2x}.

  1. Express tan⁡2x\tan 2x in terms of tan⁡x\tan x. Using the double-angle formula again:

tan⁡2x=2tan⁡x1−tan⁡2x.\tan 2x = \frac{2\tan x}{1 - \tan^2 x}.

  1. Substitute tan⁡2x\tan 2x into the expression for tan⁡4x\tan 4x. We get

tan⁡4x=2⋅2tan⁡x1−tan⁡2x1−(2tan⁡x1−tan⁡2x)2.\tan 4x = \frac{2 \cdot \frac{2\tan x}{1 - \tan^2 x}}{1 - \left(\frac{2\tan x}{1 - \tan^2 x}\right)^2}.

  1. Simplify the numerator. The numerator becomes

4tan⁡x1−tan⁡2x.\frac{4\tan x}{1 - \tan^2 x}.

  1. Simplify the denominator. The denominator is

1−4tan⁡2x(1−tan⁡2x)2.1 - \frac{4\tan^2 x}{(1 - \tan^2 x)^2}.

Write 11 as (1−tan⁡2x)2(1−tan⁡2x)2\frac{(1 - \tan^2 x)^2}{(1 - \tan^2 x)^2} to combine:

(1−tan⁡2x)2−4tan⁡2x(1−tan⁡2x)2.\frac{(1 - \tan^2 x)^2 - 4\tan^2 x}{(1 - \tan^2 x)^2}.

  1. Expand and simplify the numerator of the denominator.

(1−tan⁡2x)2−4tan⁡2x=1−2tan⁡2x+tan⁡4x−4tan⁡2x=1−6tan⁡2x+tan⁡4x.(1 - \tan^2 x)^2 - 4\tan^2 x = 1 - 2\tan^2 x + \tan^4 x - 4\tan^2 x = 1 - 6\tan^2 x + \tan^4 x.

So the denominator becomes

1−6tan⁡2x+tan⁡4x(1−tan⁡2x)2.\frac{1 - 6\tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}.

  1. Combine numerator and denominator. We now have tan⁡4x=4tan⁡x1−tan⁡2x1−6tan⁡2x+tan⁡4x(1−tan⁡2x)2.\tan 4x = \frac{\frac{4\tan x}{1 - \tan^2 x}}{\frac{1 - 6\tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}}. …

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