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

NDA 2025 · Mathematics · Trigonometric Ratios & Identities

For the following two (02) items: Let A + B = p and A + B = q ....

For the following two (02) items:
Let sin⁡A+sin⁡B=p\sin A + \sin B = p and cos⁡A+cos⁡B=q\cos A + \cos B = q.

What is p2−q2p2+q2\dfrac{p^2 - q^2}{p^2 + q^2} equal to?

Options

  1. Acos⁡(A+B)\cos(A + B)
  2. Bcos⁡(A−B)\cos(A - B)
  3. Ccos⁡(π2−A−B)\cos \left(\dfrac{\pi}{2} -A - B\right)
  4. Dcos⁡(π−A−B)\cos(\pi - A - B)

Correct answer

D. cos⁡(π−A−B)\cos(\pi - A - B)

Step-by-step solution

Given p=sin⁡A+sin⁡Bp = \sin A + \sin B and q=cos⁡A+cos⁡Bq = \cos A + \cos B.

Squaring both equations:

p2=sin⁡2A+sin⁡2B+2sin⁡Asin⁡Bp^2 = \sin^2 A + \sin^2 B + 2\sin A \sin B

q2=cos⁡2A+cos⁡2B+2cos⁡Acos⁡Bq^2 = \cos^2 A + \cos^2 B + 2\cos A \cos B

Adding the two equations:

p2+q2=(sin⁡2A+cos⁡2A)+(sin⁡2B+cos⁡2B)+2(cos⁡Acos⁡B+sin⁡Asin⁡B)p^2 + q^2 = (\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\cos A \cos B + \sin A \sin B)

p2+q2=1+1+2cos⁡(A−B)=2(1+cos⁡(A−B))p^2 + q^2 = 1 + 1 + 2\cos(A - B) = 2(1 + \cos(A - B))

Subtracting q2q^2 from p2p^2:

p2−q2=(sin⁡2A−cos⁡2A)+(sin⁡2B−cos⁡2B)−2(cos⁡Acos⁡B−sin⁡Asin⁡B)p^2 - q^2 = (\sin^2 A - \cos^2 A) + (\sin^2 B - \cos^2 B) - 2(\cos A \cos B - \sin A \sin B)

p2−q2=−cos⁡2A−cos⁡2B−2cos⁡(A+B)p^2 - q^2 = -\cos 2A - \cos 2B - 2\cos(A + B)

Using the sum-to-product formula cos⁡2A+cos⁡2B=2cos⁡(A+B)cos⁡(A−B)\cos 2A + \cos 2B = 2\cos(A + B)\cos(A - B):

p2−q2=−2cos⁡(A+B)cos⁡(A−B)−2cos⁡(A+B)p^2 - q^2 = -2\cos(A + B)\cos(A - B) - 2\cos(A + B)

p2−q2=−2cos⁡(A+B)(1+cos⁡(A−B))p^2 - q^2 = -2\cos(A + B)(1 + \cos(A - B))

Dividing p2−q2p^2 - q^2 by p2+q2p^2 + q^2:

p2−q2p2+q2=−2cos⁡(A+B)(1+cos⁡(A−B))2(1+cos⁡(A−B))\dfrac{p^2 - q^2}{p^2 + q^2} = \dfrac{-2\cos(A + B)(1 + \cos(A - B))}{2(1 + \cos(A - B))}

p2−q2p2+q2=−cos⁡(A+B)\dfrac{p^2 - q^2}{p^2 + q^2} = -\cos(A + B)

Using the property cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta:

−cos⁡(A+B)=cos⁡(π−(A+B))=cos⁡(π−A−B)-\cos(A + B) = \cos(\pi - (A + B)) = \cos(\pi - A - B)

Answer: cos⁡(π−A−B)\cos(\pi - A - B)
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