CDS 2026 (I) Elementary Mathematics, question 95

Topic: Trigonometry

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Passage · Q95 to Q96

psin⁡2α+qcos⁡2α=mp\sin^{2}\alpha+q\cos^{2}\alpha=m and

pcos⁡2β+qsin⁡2β=np\cos^{2}\beta+q\sin^{2}\beta=n

Q95

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

  1. (a) m−qp−m\frac{m-q}{p-m}Correct answer
  2. (b) m−pq−m\frac{m-p}{q-m}
  3. (c) m−qm−p\frac{m-q}{m-p}
  4. (d) m−pm−q\frac{m-p}{m-q}

Solution.

Answer: (a) m−qp−m\frac{m-q}{p-m}

  1. Divide psin⁡2α+qcos⁡2α=mp\sin^{2}\alpha + q\cos^{2}\alpha = m by cos⁡2α\cos^{2}\alpha and use sec⁡2α=1+tan⁡2α\sec^{2}\alpha = 1 + \tan^{2}\alpha.
  2. Then ptan⁡2α+q=m(1+tan⁡2α)p\tan^{2}\alpha + q = m\left( 1 + \tan^{2}\alpha \right).
  3. Thus (p−m)tan⁡2α=m−q(p - m)\tan^{2}\alpha = m - q, giving tan⁡2α=(m−q)/(p−m)\tan^{2}\alpha = (m - q)/(p - m).
  4. If p=q=mp = q = m, the condition does not determine α\alpha.