UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Use trigonometric identities and the angle sum property of a triangle to derive that the product of cosines is zero, indicating a right angle.Explanation:Given: $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$.Cross-multiply: $\sin^2 A + \sin^2 B + \sin^2 C = 2(\cos^2 A + \cos^2 B + \cos^2 C)$.Using $\sin^2\theta + \cos^2\theta = 1$, replace each $\cos^2\theta$ with $1 - \sin^2\theta$:$\sin^2 A + \sin^2 B + \sin^2 C = 2[(1 - \sin^2 A)+(1 - \sin^2 B)+(1 - \sin^2 C)] = 6 - 2(\sin^2 A + \sin^2 B + \sin^2 C)$.Let $S = \sin^2 A + \sin^2 B + \sin^2 C$. Then $S = 6 - 2S \Rightarrow 3S = 6 \Rightarrow S = 2$.Thus $\sin^2 A + \sin^2 B + \sin^2 C = 2$.Now use $\sin^2\theta = 1 - \cos^2\theta$: $(1 - \cos^2 A)+(1 - \cos^2 B)+(1 - \cos^2 C) = 2 \Rightarrow 3 - (\cos^2 A+\cos^2 B+\cos^2 C) = 2$.Hence $\cos^2 A + \cos^2 B + \cos^2 C = 1$.Rearrange: $\cos^2 A + \cos^2 B = 1 - \cos^2 C = \sin^2 C$.So $\cos^2 A + \cos^2 B - \sin^2 C = 0$.Use identity $\cos^2 X - \sin^2 Y = \cos(X+Y)\cos(X-Y)$. Then $\cos^2 A - \sin^2 C = \cos(A+C)\cos(A-C)$.Thus $\cos(A+C)\cos(A-C) + \cos^2 B = 0$.Since $A+B+C = \pi$, $A+C = \pi - B$, and $\cos(\pi - B) = -\cos B$.Substitute: $-\cos B \cos(A-C) + \cos^2 B = 0 \Rightarrow \cos B[\cos B - \cos(A-C)] = 0$.Now $\cos B - \cos(A-C) = -2\sin\left(\frac{B+A-C}{2}\right)\sin\left(\frac{B-A+C}{2}\right)$.Using $A+B+C=\pi$: $\frac{B+A-C}{2} = \frac{\pi - 2C}{2} = \frac{\pi}{2} - C$ and $\frac{B-A+C}{2} = \frac{\pi - 2A}{2} = \frac{\pi}{2} - A$.Hence $\cos B - \cos(A-C) = -2\sin(\frac{\pi}{2} - C)\sin(\frac{\pi}{2} - A) = -2\cos C \cos A$.Therefore $\cos B[-2\cos A \cos C] = 0 \Rightarrow -2\cos A \cos B \cos C = 0 \Rightarrow \cos A \cos B \cos C = 0$.At least one of $\cos A$, $\cos B$, $\cos C$ is zero, meaning that angle is $\frac{\pi}{2}$ (90°).So the triangle is right‑angled.Answer:A. right-angled