CDS 2026 (I) Elementary Mathematics, question 98

Topic: Trigonometry

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Passage · Q97 to Q98

p+qcot⁡θ=3cosec⁡θp+q\cot \theta=3\operatorname{cosec} \theta and

q−pcot⁡θ=2cosec⁡θq-p\cot \theta=2\operatorname{cosec} \theta

Q98

What is tan⁡θ\tan \theta equal to ?

  1. (a) (2q+3p)(3q−2p)\frac{(2q+3p)}{(3q-2p)}Correct answer
  2. (b) (2q−3p)(3q+2p)\frac{(2q-3p)}{(3q+2p)}
  3. (c) (3q+2p)(2q−3p)\frac{(3q+2p)}{(2q-3p)}
  4. (d) (3q−2p)(2q+3p)\frac{(3q-2p)}{(2q+3p)}

Solution.

Answer: (a) (2q+3p)(3q−2p)\frac{(2q+3p)}{(3q-2p)}

  1. Multiply both given equations by sin⁡θ\sin\theta: psin⁡θ+qcos⁡θ=3p\sin\theta + q\cos\theta = 3 and qsin⁡θ−pcos⁡θ=2q\sin\theta - p\cos\theta = 2.
  2. p times the first plus q times the second: (p2+q2)sin⁡θ=3p+2q\left( p^{2} + q^{2} \right)\sin\theta = 3p + 2q.
  3. q times the first minus p times the second: (p2+q2)cos⁡θ=3q−2p\left( p^{2} + q^{2} \right)\cos\theta = 3q - 2p.
  4. Divide: tan⁡θ=3p+2q3q−2p\tan\theta = \frac{3p + 2q}{3q - 2p}, which is (a) with the numerator written as 2q+3p2q + 3p.

Trap: (a) prints the numerator in a different order from the working, and (d) is the same fraction upside down, which is cot⁡θ\cot\theta.