CDS 2026 (I) Elementary Mathematics, question 99

Topic: Trigonometry

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Passage · Q99 to Q100

p=sin⁡θ1+cos⁡θ+sin⁡θp=\frac{\sin \theta}{1+\cos \theta+\sin \theta} and

q=1+sin⁡θ1+sin⁡θ−cos⁡θq=\frac{1+\sin \theta}{1+\sin \theta-\cos \theta}

Q99

Which one of the following is correct ?

  1. (a) p−q=0p-q=0
  2. (b) 2pq−1=02pq-1=0Correct answer
  3. (c) pq−2=0pq-2=0
  4. (d) pq−1=0pq-1=0

Solution.

Answer: (b) 2pq−1=02pq-1=0

  1. Let s=sin⁡θs = \sin\theta and c=cos⁡θc = \cos\theta.
  2. Multiplying the definitions gives pq=s(1+s)/[(1+s+c)(1+s−c)]pq = s(1 + s)/\left\lbrack (1 + s + c)(1 + s - c) \right\rbrack.
  3. The denominator is (1+s)2−c2=1+2s+s2−(1−s2)=2s(1+s)(1 + s)^{2} - c^{2} = 1 + 2s + s^{2} - \left( 1 - s^{2} \right) = 2s(1 + s).
  4. Hence pq=1/2pq = 1/2 wherever the original fractions are defined.