Q176 / 270 · NDA 2025 (Phase 2) · Mathematics

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

What is $\dfrac{p^2 - q^2}{p^2 + q^2}$ equal to?

  • A
    $\cos(A + B)$
  • B
    $\cos(A - B)$
  • C
    $\cos\left(\dfrac{\pi}{2} - A - B\right)$
  • Correct answer
    D
    $\cos(\pi - A - B)$

Solution

Given $p = \sin A + \sin B$ and $q = \cos A + \cos B$.

Squaring both equations:

$p^2 = \sin^2 A + \sin^2 B + 2\sin A \sin B$

$q^2 = \cos^2 A + \cos^2 B + 2\cos A \cos B$

Adding the two equations:

$p^2 + q^2 = (\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\cos A \cos B + \sin A \sin B)$

$p^2 + q^2 = 1 + 1 + 2\cos(A - B) = 2(1 + \cos(A - B))$

Subtracting $q^2$ from $p^2$:

$p^2 - q^2 = (\sin^2 A - \cos^2 A) + (\sin^2 B - \cos^2 B) - 2(\cos A \cos B - \sin A \sin B)$

$p^2 - q^2 = -\cos 2A - \cos 2B - 2\cos(A + B)$

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

$p^2 - q^2 = -2\cos(A + B)\cos(A - B) - 2\cos(A + B)$

$p^2 - q^2 = -2\cos(A + B)(1 + \cos(A - B))$

Dividing $p^2 - q^2$ by $p^2 + q^2$:

$\dfrac{p^2 - q^2}{p^2 + q^2} = \dfrac{-2\cos(A + B)(1 + \cos(A - B))}{2(1 + \cos(A - B))}$

$\dfrac{p^2 - q^2}{p^2 + q^2} = -\cos(A + B)$

Using the property $\cos(\pi - \theta) = -\cos \theta$:

$-\cos(A + B) = \cos(\pi - (A + B)) = \cos(\pi - A - B)$

Answer: $\cos(\pi - A - B)$

Hence, the correct answer is option: 'D'.

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