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NDA 1 2025 Maths, question 43

Topic: Trigonometry

Passage · Q41 to Q43

Let p=tan⁡2α−tan⁡αp = \tan 2\alpha - \tan\alpha and q=cot⁡α−cot⁡2αq = \cot\alpha - \cot 2\alpha.

Q43

What is tan⁡2α\tan^{2}\alpha equal to?

  1. (a) pqp+q\frac{pq}{p + q}
  2. (b) p+2qp\frac{p + 2q}{p}
  3. (c) pp+2q\frac{p}{p + 2q}Correct answer
  4. (d) p2p+q\frac{p}{2p + q}

Solution.

Answer: (c) pp+2q\frac{p}{p + 2q}

  1. Let t=tan⁡αt = \tan\alpha.
  2. The double-angle formula gives p=t(1+t2)1−t2p = \frac{t\left( 1 + t^{2} \right)}{1 - t^{2}} and q=1+t22tq = \frac{1 + t^{2}}{2t}.
  3. Hence p/q=2t2/(1−t2)p/q = 2t^{2}/\left( 1 - t^{2} \right).
  4. Rearranging, p=(p+2q)t2p = (p + 2q)t^{2}, so tan⁡2α=p/(p+2q)\tan^{2}\alpha = p/(p + 2q).