CDS 2026 (I) Elementary Mathematics, question 42

Topic: Trigonometry

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Q42

For 0<θ<π20<\theta<\frac{\pi}{2}, consider the following :

I. (tan⁡4θ+tan⁡6θ)(cot⁡4θ+cot⁡6θ)=sec⁡2θcosec⁡2θ(\tan^{4}\theta+\tan^{6}\theta)(\cot^{4}\theta+\cot^{6}\theta)=\sec^{2}\theta\operatorname{cosec}^{2}\theta

II. tan⁡θ+sin⁡θtan⁡θ−sin⁡θ=cot⁡2θ(sec⁡θ+1)2\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\cot^{2}\theta(\sec \theta+1)^{2}

Which of the above is/are identities ?

  1. (a) I only
  2. (b) II only
  3. (c) Both I and IICorrect answer
  4. (d) Neither I nor II

Solution.

Answer: (c) Both I and II

  1. For I, factor the left side as tan⁡4θsec⁡2θ×cot⁡4θcosec2θ\tan^{4}\theta\sec^{2}\theta \times \cot^{4}\theta{cosec}^{2}\theta.
  2. Since tan⁡θcot⁡θ=1\tan\theta\cot\theta = 1, this reduces to sec⁡2θcosec2θ\sec^{2}\theta{cosec}^{2}\theta.
  3. For II, dividing numerator and denominator by sin⁡θ\sin\theta gives (sec⁡θ+1)/(sec⁡θ−1)\left( \sec\theta + 1 \right)/\left( \sec\theta - 1 \right).
  4. Multiply by (sec⁡θ+1)/(sec⁡θ+1)\left( \sec\theta + 1 \right)/\left( \sec\theta + 1 \right), then use sec⁡2θ−1=tan⁡2θ\sec^{2}\theta - 1 = \tan^{2}\theta to obtain cot⁡2θ(sec⁡θ+1)2\cot^{2}\theta\left( \sec\theta + 1 \right)^{2}.
  5. The acute-angle condition keeps the denominators nonzero.