UPSC CDS 2 2021 Math Paper

Concept:Use trigonometric identities to simplify the given equation, solve for $\cos\theta$, and then compute the required expression.Explanation:Given: $\frac{\cos^{2}\theta - 3\cos\theta + 2}{\sin^{2}\theta} = 1$, with $0^{\circ} < \theta < 90^{\circ}$.Step 1: Multiply both sides by $\sin^{2}\theta$:$\cos^{2}\theta - 3\cos\theta + 2 = \sin^{2}\theta$.Step 2: Use the identity $\sin^{2}\theta = 1 - \cos^{2}\theta$:$\cos^{2}\theta - 3\cos\theta + 2 = 1 - \cos^{2}\theta$.Step 3: Bring all terms to one side:$\cos^{2}\theta - 3\cos\theta + 2 - 1 + \cos^{2}\theta = 0$⇒ $2\cos^{2}\theta - 3\cos\theta + 1 = 0$.Step 4: Factor the quadratic in $\cos\theta$:$2\cos^{2}\theta - 2\cos\theta - \cos\theta + 1 = 0$⇒ $2\cos\theta(\cos\theta - 1) - 1(\cos\theta - 1) = 0$⇒ $(\cos\theta - 1)(2\cos\theta - 1) = 0$.Step 5: Solve for $\cos\theta$:$\cos\theta = 1$ or $\cos\theta = \frac{1}{2}$.Since $0^{\circ} < \theta < 90^{\circ}$, $\cos\theta = 1$ gives $\theta = 0^{\circ}$ (not allowed).Thus $\cos\theta = \frac{1}{2}$ ⇒ $\theta = 60^{\circ}$.Step 6: Compute $\sin^{2}\theta + \cos\theta$ at $\theta = 60^{\circ}$:$\sin^{2}60^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4}$$\cos 60^{\circ} = \frac{1}{2}$Sum = $\frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$.Answer:$\frac{5}{4}$ (Option A).