CDS 2026 (I) Elementary Mathematics, question 96

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

Q96

What is cot⁡2β\cot^{2}\beta equal to ?

  1. (a) n−qp−n\frac{n-q}{p-n}Correct answer
  2. (b) n−pq−n\frac{n-p}{q-n}
  3. (c) n−pn−q\frac{n-p}{n-q}
  4. (d) n−qn−p\frac{n-q}{n-p}

Solution.

Answer: (a) n−qp−n\frac{n-q}{p-n}

  1. Divide pcos⁡2β+qsin⁡2β=np\cos^{2}\beta + q\sin^{2}\beta = n by sin⁡2β\sin^{2}\beta and use cosec2β=1+cot⁡2β{cosec}^{2}\beta = 1 + \cot^{2}\beta.
  2. Then pcot⁡2β+q=n(1+cot⁡2β)p\cot^{2}\beta + q = n\left( 1 + \cot^{2}\beta \right).
  3. Thus (p−n)cot⁡2β=n−q(p - n)\cot^{2}\beta = n - q, giving cot⁡2β=(n−q)/(p−n)\cot^{2}\beta = (n - q)/(p - n).
  4. The degenerate case p=q=np = q = n does not determine β\beta.