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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 (p+q)(p + q) equal to?

Options

  1. Asec⁡4α\sec 4\alpha
  2. Bcsc⁡4α\csc 4\alpha
  3. C2sec⁡4α2\sec 4\alpha
  4. D2csc⁡4α2\csc 4\alpha

Correct answer

D. 2csc⁡4α2\csc 4\alpha

Step-by-step solution

Given p=tan⁡2α−tan⁡αp = \tan 2\alpha - \tan\alpha and q=cot⁡α−cot⁡2αq = \cot\alpha - \cot 2\alpha.

Simplifying pp:
p=sin⁡2αcos⁡2α−sin⁡αcos⁡α=sin⁡2αcos⁡α−cos⁡2αsin⁡αcos⁡2αcos⁡α=sin⁡(2α−α)cos⁡2αcos⁡α=sin⁡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} = \dfrac{\sin(2\alpha - \alpha)}{\cos 2\alpha \cos\alpha} = \dfrac{\sin\alpha}{\cos 2\alpha \cos\alpha}

Simplifying qq:
q=cos⁡αsin⁡α−cos⁡2αsin⁡2α=cos⁡αsin⁡2α−sin⁡αcos⁡2αsin⁡αsin⁡2α=sin⁡(2α−α)sin⁡αsin⁡2α=sin⁡αsin⁡αsin⁡2α=1sin⁡2αq = \dfrac{\cos\alpha}{\sin\alpha} - \dfrac{\cos 2\alpha}{\sin 2\alpha} = \dfrac{\cos\alpha \sin 2\alpha - \sin\alpha \cos 2\alpha}{\sin\alpha \sin 2\alpha} = \dfrac{\sin(2\alpha - \alpha)}{\sin\alpha \sin 2\alpha} = \dfrac{\sin\alpha}{\sin\alpha \sin 2\alpha} = \dfrac{1}{\sin 2\alpha}

Now, adding pp and qq:
p+q=sin⁡αcos⁡2αcos⁡α+1sin⁡2αp + q = \dfrac{\sin\alpha}{\cos 2\alpha \cos\alpha} + \dfrac{1}{\sin 2\alpha}

Using sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha = 2\sin\alpha \cos\alpha:
p+q=sin⁡αcos⁡2αcos⁡α+12sin⁡αcos⁡αp + q = \dfrac{\sin\alpha}{\cos 2\alpha \cos\alpha} + \dfrac{1}{2\sin\alpha \cos\alpha}

p+q=2sin⁡2α+cos⁡2α2sin⁡αcos⁡αcos⁡2αp + q = \dfrac{2\sin^2\alpha + \cos 2\alpha}{2\sin\alpha \cos\alpha \cos 2\alpha}

Using the identity cos⁡2α=1−2sin⁡2α\cos 2\alpha = 1 - 2\sin^2\alpha in the numerator and 2sin⁡αcos⁡α=sin⁡2α2\sin\alpha \cos\alpha = \sin 2\alpha in the denominator:
p+q=2sin⁡2α+(1−2sin⁡2α)sin⁡2αcos⁡2αp + q = \dfrac{2\sin^2\alpha + (1 - 2\sin^2\alpha)}{\sin 2\alpha \cos 2\alpha}

p+q=1sin⁡2αcos⁡2αp + q = \dfrac{1}{\sin 2\alpha \cos 2\alpha}

Multiplying and dividing by 22:
p+q=22sin⁡2αcos⁡2α=2sin⁡4αp + q = \dfrac{2}{2\sin 2\alpha \cos 2\alpha} = \dfrac{2}{\sin 4\alpha}

p+q=2csc⁡4αp + q = 2\csc 4\alpha

Answer: 2csc⁡4α2\csc 4\alpha
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