CDS 2026 (I) Elementary Mathematics, question 94

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}

Q94

What is p4q2+p2q4p^{4}q^{2}+p^{2}q^{4} equal to ?

  1. (a) −2-2
  2. (b) −1-1
  3. (c) 00
  4. (d) 11Correct answer

Solution.

Answer: (d) 11

  1. Multiplying p3=cos⁡2θ/sin⁡θp^{3} = \cos^{2}\theta/\sin\theta and q3=sin⁡2θ/cos⁡θq^{3} = \sin^{2}\theta/\cos\theta gives p3q3=sin⁡θcos⁡θp^{3}q^{3} = \sin\theta\cos\theta.
  2. Now p6q3=(cos⁡2θ/sin⁡θ)(sin⁡θcos⁡θ)=cos⁡3θp^{6}q^{3} = \left( \cos^{2}\theta/\sin\theta \right)\left( \sin\theta\cos\theta \right) = \cos^{3}\theta, so p2q=cos⁡θp^{2}q = \cos\theta.
  3. Similarly pq2=sin⁡θpq^{2} = \sin\theta.
  4. Squaring and adding gives p4q2+p2q4=cos⁡2θ+sin⁡2θ=1p^{4}q^{2} + p^{2}q^{4} = \cos^{2}\theta + \sin^{2}\theta = 1.