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NCERT Exemplar · Q32

Q.If tan⁡θ=12\tan\theta = \dfrac{1}{2} and tan⁡ϕ=13\tan\phi = \dfrac{1}{3}, then the value of θ+ϕ\theta + \phi is
(A) π6\dfrac{\pi}{6}
(B) π\pi
(C) 00
(D) π4\dfrac{\pi}{4}

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Using the tangent addition formula, tan⁡(θ+ϕ)=tan⁡θ+tan⁡ϕ1−tan⁡θtan⁡ϕ=1/2+1/31−(1/2)(1/3)=1\tan(\theta+\phi) = \frac{\tan\theta + \tan\phi}{1 - \tan\theta \tan\phi} = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = 1, so θ+ϕ=π4\theta+\phi = \frac{\pi}{4} (option D).

The core idea here is that when you know the tangents of two angles individually, the tangent of their sum is given by a clean formula. Instead of trying to find each angle separately (which would involve messy inverse trig values), we combine them directly.

Why this works: The tangent addition formula is derived from sine and cosine addition formulas, and it’s the most efficient tool when both tangents are known rational numbers. Here, the numbers 12\frac12 and 13\frac13 are chosen so that the denominator becomes 1−16=561 - \frac16 = \frac56, and the numerator becomes 56\frac56, giving a perfect 11. That’s a strong hint that the sum is a standard angle.

Let’s go step by step.

  1. Recall the tangent addition formula For any two angles θ\theta and ϕ\phi (where cos⁡θ≠0\cos\theta \neq 0, cos⁡ϕ≠0\cos\phi \neq 0, and tan⁡θtan⁡ϕ≠1\tan\theta\tan\phi \neq 1),

tan⁡(θ+ϕ)=tan⁡θ+tan⁡ϕ1−tan⁡θtan⁡ϕ.\tan(\theta + \phi) = \frac{\tan\theta + \tan\phi}{1 - \tan\theta \tan\phi}.

  1. Substitute the given values We have tan⁡θ=12\tan\theta = \frac12 and tan⁡ϕ=13\tan\phi = \frac13. So

tan⁡(θ+ϕ)=12+131−(12)(13).\tan(\theta + \phi) = \frac{\frac12 + \frac13}{1 - \left(\frac12\right)\left(\frac13\right)}.

  1. Simplify numerator and denominator Numerator: 12+13=36+26=56\frac12 + \frac13 = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}. Denominator: 1−16=66−16=561 - \frac16 = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}. So

tan⁡(θ+ϕ)=5/65/6=1.\tan(\theta + \phi) = \frac{5/6}{5/6} = 1.

  1. Interpret the result …

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