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NCERT Exemplar · Q74

Q.If tan⁡θ+tan⁡2θ+3 tan⁡θ tan⁡2θ=3\tan\theta + \tan 2\theta + \sqrt{3}\,\tan\theta\,\tan 2\theta = \sqrt{3}, then θ=nπ3+π9\theta = \dfrac{n\pi}{3} + \dfrac{\pi}{9}.

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The key idea is to rewrite the given equation using the tangent addition formula for tan⁡(3θ)=tan⁡(θ+2θ)\tan(3\theta) = \tan(\theta + 2\theta). This transforms the expression into tan⁡(3θ)=3\tan(3\theta) = \sqrt{3}, leading to the general solution θ=nπ3+π9\theta = \frac{n\pi}{3} + \frac{\pi}{9}.

This problem is a classic example of how a seemingly messy trigonometric equation can be tamed by recognizing a hidden identity. The expression tan⁡θ+tan⁡2θ+3 tan⁡θ tan⁡2θ\tan\theta + \tan 2\theta + \sqrt{3}\,\tan\theta\,\tan 2\theta looks like it came straight from the formula for tan⁡(A+B)\tan(A+B) — but with a twist. Let’s see why.

Recall the tangent addition formula:

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B.\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}.

If we rearrange this, we get:

tan⁡A+tan⁡B=tan⁡(A+B)⋅(1−tan⁡Atan⁡B).\tan A + \tan B = \tan(A+B) \cdot (1 - \tan A \tan B).

Now, if we add a term like 3 tan⁡Atan⁡B\sqrt{3}\,\tan A \tan B to both sides, we might be able to match the given equation. Here, A=θA = \theta and B=2θB = 2\theta, so A+B=3θA+B = 3\theta. The number 3\sqrt{3} is no accident — it’s tan⁡60∘=tan⁡π3\tan 60^\circ = \tan \frac{\pi}{3}. So the equation is hinting that tan⁡(3θ)=3\tan(3\theta) = \sqrt{3}.

Let’s work through it step by step.

  1. Start with the given equation:

tan⁡θ+tan⁡2θ+3 tan⁡θ tan⁡2θ=3.\tan\theta + \tan 2\theta + \sqrt{3}\,\tan\theta\,\tan 2\theta = \sqrt{3}.

  1. Isolate the sum of tangents: Move the product term to the right side:

tan⁡θ+tan⁡2θ=3−3 tan⁡θ tan⁡2θ.\tan\theta + \tan 2\theta = \sqrt{3} - \sqrt{3}\,\tan\theta\,\tan 2\theta.

Factor 3\sqrt{3} on the right:

tan⁡θ+tan⁡2θ=3 (1−tan⁡θ tan⁡2θ).\tan\theta + \tan 2\theta = \sqrt{3}\,(1 - \tan\theta\,\tan 2\theta).

  1. Recognize the tangent addition formula: Compare with tan⁡(θ+2θ)=tan⁡θ+tan⁡2θ1−tan⁡θ tan⁡2θ\tan(\theta + 2\theta) = \frac{\tan\theta + \tan 2\theta}{1 - \tan\theta\,\tan 2\theta}. If 1−tan⁡θ tan⁡2θ≠01 - \tan\theta\,\tan 2\theta \neq 0, we can divide both sides of the equation by it:

tan⁡θ+tan⁡2θ1−tan⁡θ tan⁡2θ=3.\frac{\tan\theta + \tan 2\theta}{1 - \tan\theta\,\tan 2\theta} = \sqrt{3}.

The left side is exactly tan⁡(3θ)\tan(3\theta). …

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