CDS 2026 (I) Elementary Mathematics, question 93

Topic: Trigonometry

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Passage · Q93 to Q94

cosec⁡θ−sin⁡θ=p3\operatorname{cosec} \theta-\sin \theta=p^{3} and sec⁡θ−cos⁡θ=q3\sec \theta-\cos \theta=q^{3}

Q93

What is tan⁡θ\tan \theta equal to ?

  1. (a) pq\frac{p}{q}
  2. (b) qp\frac{q}{p}Correct answer
  3. (c) pqpq
  4. (d) pq\sqrt{pq}

Solution.

Answer: (b) qp\frac{q}{p}

  1. Rewrite the shared equations as p3=cos⁡2θ/sin⁡θp^{3} = \cos^{2}\theta/\sin\theta and q3=sin⁡2θ/cos⁡θq^{3} = \sin^{2}\theta/\cos\theta.
  2. Dividing gives q3/p3=sin⁡3θ/cos⁡3θ=tan⁡3θq^{3}/p^{3} = \sin^{3}\theta/\cos^{3}\theta = \tan^{3}\theta.
  3. Taking real cube roots gives tan⁡θ=q/p\tan\theta = q/p.