CDS 2026 (I) Elementary Mathematics, question 91

Topic: Trigonometry

Answers by SolvedPaper. UPSC has not released the official answer key for this paper yet. We will update it when they do.

Passage · Q91 to Q92

Let 12(tan⁡θ+cot⁡θ)=2512(\tan \theta+\cot \theta)=25, where 45∘<θ<90∘45^{\circ}<\theta<90^{\circ}.

Q91

What is a value of (sin⁡θ−cos⁡θ)(\sin \theta-\cos \theta) ?

  1. (a) −15-\frac{1}{5}
  2. (b) −25-\frac{2}{5}
  3. (c) 15\frac{1}{5}Correct answer
  4. (d) 25\frac{2}{5}

Solution.

Answer: (c) 15\frac{1}{5}

  1. Use tan⁡θ+cot⁡θ=1/(sin⁡θcos⁡θ)\tan\theta + \cot\theta = 1/\left( \sin\theta\cos\theta \right).
  2. The shared condition gives sin⁡θcos⁡θ=12/25\sin\theta\cos\theta = 12/25.
  3. Then (sin⁡θ−cos⁡θ)2=1−2(12/25)=1/25\left( \sin\theta - \cos\theta \right)^{2} = 1 - 2(12/25) = 1/25.
  4. Since 45∘<θ<90∘,sin⁡θ>cos⁡θ45^{\circ} < \theta < 90^{\circ},\sin\theta > \cos\theta, so take the positive root.