UPSC CDS 1 2022 Math Paper

Concept:The given equation $\cos\theta + \sec\theta - 2 = 0$ can be rewritten as $x + \frac{1}{x} = 2$, where $x = \cos\theta$.Since $0 \le \theta < \frac{\pi}{2}$, $\cos\theta$ is positive and its maximum is $1$ at $\theta = 0^{\circ}$.The equation $x + \frac{1}{x} = 2$ holds only when $x = 1$.Explanation:Set $x = \cos\theta$. Then $\sec\theta = \frac{1}{x}$ because $\sec\theta = \frac{1}{\cos\theta}$.The given condition becomes $x + \frac{1}{x} - 2 = 0$, i.e., $x + \frac{1}{x} = 2$.Multiply both sides by $x$: $x^2 + 1 = 2x \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2 = 0$.Thus $x = 1$, so $\cos\theta = 1$ (and $\sec\theta = 1$).Now compute the required expression:$\cos^4\theta + \sec^4\theta - 2 = (1)^4 + (1)^4 - 2 = 1 + 1 - 2 = 0$.Alternatively, starting from $x + \frac{1}{x} = 2$, square both sides: $x^2 + \frac{1}{x^2} + 2 = 4 \Rightarrow x^2 + \frac{1}{x^2} = 2$.Square again: $x^4 + \frac{1}{x^4} + 2 = 4 \Rightarrow x^4 + \frac{1}{x^4} = 2$.Therefore $x^4 + \frac{1}{x^4} - 2 = 0$, i.e., $\cos^4\theta + \sec^4\theta - 2 = 0$.Answer:$0$ (Option C).