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NDA 1 2025 Maths, question 32

Topic: Trigonometry

Passage · Q31 to Q33

Let p=sin⁡35∘p = \sin 35^{\circ}, q=sin⁡25∘q = \sin 25^{\circ} and r=sin⁡(−95∘)r = \sin\left( - 95^{\circ} \right).

Q32

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

  1. (a) −3/4- 3/4Correct answer
  2. (b) 0
  3. (c) 1/41/4
  4. (d) 3/43/4

Solution.

Answer: (a) −3/4- 3/4

  1. As p+q=cos⁡5∘p + q = \cos 5^{\circ} and r=−cos⁡5∘r = - \cos 5^{\circ}, we have pq+qr+rp=pq−cos⁡2(5∘)pq + qr + rp = pq - \cos^{2}\left( 5^{\circ} \right).
  2. Now pq=cos⁡10∘−cos⁡60∘2=12cos⁡10∘−14pq = \frac{\cos 10^{\circ} - \cos 60^{\circ}}{2} = \frac{1}{2}\cos 10^{\circ} - \frac{1}{4} and cos⁡2(5∘)=1+cos⁡10∘2\cos^{2}\left( 5^{\circ} \right) = \frac{1 + \cos 10^{\circ}}{2}.
  3. Subtraction gives −3/4- 3/4.