Q181 / 269 · NDA 2025 (Phase 1) · Mathematics

Consider the following for the three (03) items that follow:
Let $p = \sin 35°$, $q = \sin 25°$ and $r = \sin(-95°)$.

What is $(pq + qr + rp)$ equal to?

  • Correct answer
    A
    $-\dfrac{3}{4}$
  • B
    $0$
  • C
    $\dfrac{1}{4}$
  • D
    $\dfrac{3}{4}$

Solution

Given $p = \sin 35^\circ$, $q = \sin 25^\circ$, and $r = \sin(-95^\circ) = -\sin(90^\circ + 5^\circ) = -\cos 5^\circ$.

$p + q + r = \sin 35^\circ + \sin 25^\circ - \cos 5^\circ$

Using $\sin A + \sin B = 2 \sin\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A-B}{2}\right)$:

$p + q + r = 2 \sin 30^\circ \cos 5^\circ - \cos 5^\circ = 2 \left(\dfrac{1}{2}\right) \cos 5^\circ - \cos 5^\circ = 0$

We know that $(p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) = 0$

$pq + qr + rp = -\dfrac{1}{2}(p^2 + q^2 + r^2)$

Now, $p^2 + q^2 + r^2 = \sin^2 35^\circ + \sin^2 25^\circ + \cos^2 5^\circ$

Using $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$:

$p^2 + q^2 + r^2 = \dfrac{1 - \cos 70^\circ}{2} + \dfrac{1 - \cos 50^\circ}{2} + \dfrac{1 + \cos 10^\circ}{2}$

$= \dfrac{3}{2} - \dfrac{1}{2}(\cos 70^\circ + \cos 50^\circ - \cos 10^\circ)$

Using $\cos C + \cos D = 2 \cos\left(\dfrac{C+D}{2}\right) \cos\left(\dfrac{C-D}{2}\right)$:

$\cos 70^\circ + \cos 50^\circ = 2 \cos 60^\circ \cos 10^\circ = 2 \left(\dfrac{1}{2}\right) \cos 10^\circ = \cos 10^\circ$

So, $p^2 + q^2 + r^2 = \dfrac{3}{2} - \dfrac{1}{2}(\cos 10^\circ - \cos 10^\circ) = \dfrac{3}{2}$

Therefore, $pq + qr + rp = -\dfrac{1}{2} \left(\dfrac{3}{2}\right) = -\dfrac{3}{4}$

Answer: $-\dfrac{3}{4}$

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

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