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NCERT Exemplar · Q58

Q.If tan⁡θ=ab\tan\theta = \dfrac{a}{b}, then bcos⁡2θ+asin⁡2θb\cos 2\theta + a\sin 2\theta is equal to
(A) aa
(B) bb
(C) ab\dfrac{a}{b}
(D) None

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Express cos⁡2θ\cos 2\theta and sin⁡2θ\sin 2\theta in terms of tan⁡θ\tan\theta, substitute tan⁡θ=ab\tan\theta = \frac{a}{b}, and simplify using the identity sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta. The result is bb.

The heart of this problem is recognizing that when you know tan⁡θ\tan\theta, you can express all other trigonometric functions of θ\theta (and its multiples) in terms of that ratio. The double-angle formulas for sine and cosine can be rewritten entirely in terms of tan⁡θ\tan\theta, which makes substitution straightforward.

Let me show you why this works and how to execute it cleanly.

Understanding the approach

When tan⁡θ=ab\tan\theta = \frac{a}{b}, we're essentially saying that in a right triangle with angle θ\theta, the opposite side is aa and the adjacent side is bb. Every trigonometric function of θ\theta can be derived from this single piece of information.

For double angles, the key identities are:

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

Both can be expressed in terms of tan⁡θ\tan\theta by dividing through by cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta appropriately.

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

Step-by-step solution

  1. Substitute the given value into the double-angle formulas.

    Since tan⁡θ=ab\tan\theta = \frac{a}{b}, we have:

cos⁡2θ=1−(ab)21+(ab)2=1−a2b21+a2b2=b2−a2b2+a2\cos 2\theta = \frac{1 - \left(\frac{a}{b}\right)^2}{1 + \left(\frac{a}{b}\right)^2} = \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} = \frac{b^2 - a^2}{b^2 + a^2}

  1. Similarly for sin⁡2θ\sin 2\theta: sin⁡2θ=2⋅ab1+(ab)2=2ab1+a2b2=2abb2+a2b2=2abb2+a2\sin 2\theta = \frac{2 \cdot \frac{a}{b}}{1 + \left(\frac{a}{b}\right)^2} = \frac{\frac{2a}{b}}{1 + \frac{a^2}{b^2}} = \frac{\frac{2a}{b}}{\frac{b^2 + a^2}{b^2}} = \frac{2ab}{b^2 + a^2} …

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