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NCERT Exemplar · Q35

Q.The value of 1−tan⁡215∘1+tan⁡215∘\dfrac{1 - \tan^2 15^\circ}{1 + \tan^2 15^\circ} is
(A) 11
(B) 3\sqrt{3}
(C) 32\dfrac{\sqrt{3}}{2}
(D) 22

Chhattisgarh CgbseMCQ· 1mImportance★★★★★est
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The expression 1−tan⁡215∘1+tan⁡215∘\frac{1 - \tan^2 15^\circ}{1 + \tan^2 15^\circ} simplifies using the identity cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}, giving cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}. So the answer is (C).

This problem is a classic test of your ability to connect trigonometric identities with known angle values. The expression 1−tan⁡2θ1+tan⁡2θ\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} might look unfamiliar at first, but it’s actually a disguised form of cos⁡2θ\cos 2\theta. Let’s see why.

Recall the double-angle formulas for cosine:

cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2 \theta - \sin^2 \theta

If we divide numerator and denominator by cos⁡2θ\cos^2 \theta (assuming cos⁡θ≠0\cos \theta \neq 0), we get:

cos⁡2θ=cos⁡2θ−sin⁡2θcos⁡2θ+sin⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}

because sin⁡2θcos⁡2θ=tan⁡2θ\frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta and cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1.

cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\displaystyle \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}

This identity works for any angle where tan⁡θ\tan \theta is defined. So our given expression is simply cos⁡(2×15∘)=cos⁡30∘\cos(2 \times 15^\circ) = \cos 30^\circ.

Now, cos⁡30∘\cos 30^\circ is a standard value you must know:

cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}

Watch out

A common mistake is to confuse tan⁡215∘\tan^2 15^\circ with tan⁡15∘\tan 15^\circ squared — they are the same, but students sometimes misapply the identity 1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta only when θ\theta is in degrees or radians consistently. Here, 15∘15^\circ is fine.

Let’s walk through the steps clearly: …

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