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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2025 · Mathematics · Trigonometric Ratios & Identities

Consider the following for the three (03) items that follow: Let p...

Consider the following for the three (03) items that follow:
Let p=tan⁡2α−tan⁡αp = \tan 2\alpha - \tan\alpha and q=cot⁡α−cot⁡2αq = \cot\alpha - \cot 2\alpha.

What is tan⁡2α\tan^2\alpha equal to?

Options

  1. Apqp+q\dfrac{pq}{p+q}
  2. Bp+2qp\dfrac{p+2q}{p}
  3. Cpp+2q\dfrac{p}{p+2q}
  4. Dp2p+q\dfrac{p}{2p+q}

Correct answer

C. pp+2q\dfrac{p}{p+2q}

Step-by-step solution

Given p=tan⁡2α−tan⁡αp = \tan 2\alpha - \tan\alpha

p=sin⁡2αcos⁡2α−sin⁡αcos⁡α=sin⁡2αcos⁡α−cos⁡2αsin⁡αcos⁡2αcos⁡αp = \dfrac{\sin 2\alpha}{\cos 2\alpha} - \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\sin 2\alpha \cos\alpha - \cos 2\alpha \sin\alpha}{\cos 2\alpha \cos\alpha}

p=sin⁡(2α−α)cos⁡2αcos⁡α=sin⁡αcos⁡2αcos⁡α=tan⁡αcos⁡2αp = \dfrac{\sin(2\alpha - \alpha)}{\cos 2\alpha \cos\alpha} = \dfrac{\sin\alpha}{\cos 2\alpha \cos\alpha} = \dfrac{\tan\alpha}{\cos 2\alpha}

Also given q=cot⁡α−cot⁡2αq = \cot\alpha - \cot 2\alpha

q=cos⁡αsin⁡α−cos⁡2αsin⁡2α=sin⁡2αcos⁡α−cos⁡2αsin⁡αsin⁡2αsin⁡αq = \dfrac{\cos\alpha}{\sin\alpha} - \dfrac{\cos 2\alpha}{\sin 2\alpha} = \dfrac{\sin 2\alpha \cos\alpha - \cos 2\alpha \sin\alpha}{\sin 2\alpha \sin\alpha}

q=sin⁡(2α−α)sin⁡2αsin⁡α=sin⁡αsin⁡2αsin⁡α=1sin⁡2αq = \dfrac{\sin(2\alpha - \alpha)}{\sin 2\alpha \sin\alpha} = \dfrac{\sin\alpha}{\sin 2\alpha \sin\alpha} = \dfrac{1}{\sin 2\alpha}

Dividing pp by qq, we get:

pq=tan⁡αcos⁡2α×sin⁡2α=tan⁡αtan⁡2α\dfrac{p}{q} = \dfrac{\tan\alpha}{\cos 2\alpha} \times \sin 2\alpha = \tan\alpha \tan 2\alpha

Using the formula tan⁡2α=2tan⁡α1−tan⁡2α\tan 2\alpha = \dfrac{2\tan\alpha}{1 - \tan^2\alpha}:

pq=tan⁡α(2tan⁡α1−tan⁡2α)=2tan⁡2α1−tan⁡2α\dfrac{p}{q} = \tan\alpha \left( \dfrac{2\tan\alpha}{1 - \tan^2\alpha} \right) = \dfrac{2\tan^2\alpha}{1 - \tan^2\alpha}

p(1−tan⁡2α)=2qtan⁡2αp(1 - \tan^2\alpha) = 2q\tan^2\alpha

p−ptan⁡2α=2qtan⁡2αp - p\tan^2\alpha = 2q\tan^2\alpha

p=tan⁡2α(p+2q)p = \tan^2\alpha(p + 2q)

tan⁡2α=pp+2q\tan^2\alpha = \dfrac{p}{p + 2q}

Answer: pp+2q\dfrac{p}{p+2q}
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