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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 31-33

Consider the following for the three (03) items that follow: Let p=sin35^∘, q=sin25^∘ and r=sin(−95^∘).

Question : 32 of 120

Marks: +1, -0

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

Solution:

The expression to simplify is

pq+qr+rp

, where:

p = \sin (35^{\circ})

q = \sin (25^{\circ})

r = \sin (-95^{\circ}) = -\cos (5^{\circ})

Thus, we need to evaluate:

pq+qr+rp = \sin (35^{\circ}) \sin (25^{\circ}) + \sin (25^{\circ}) (-\cos (5^{\circ})) + (-\cos (5^{\circ})) \sin (35^{\circ})

Using trigonometric identities, we simplify the expression:

pq+qr+rp = -\sin^2 (25^{\circ}) - \sin (95^{\circ}) \sin (35^{\circ})

Using the sum-to-product identity and known values for sine and cosine:

= \frac{1}{2} \left[ -2 \sin (25^{\circ}) -2 \sin (95^{\circ}) \cdot \sin (35^{\circ}) \right]

= \frac{1}{2} \left[ -1 + \cos (50^{\circ}) + \cos (130^{\circ}) - \cos (60^{\circ}) \right]

= \frac{1}{2} \left[ -1 + \cos (50^{\circ}) - \cos (50^{\circ}) - \frac{1}{2} \right]

= \frac{1}{2} \left[ -1 - \frac{1}{2} \right] = -\frac{3}{4}

\therefore

The value of

pq+qr+rp

-\frac{3}{4}

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