UPSC CDS 2 2020 Math Paper

Concept:Express the given trigonometric expressions in terms of $\sin\theta$ and $\cos\theta$, then simplify the required expression using algebraic manipulation.Explanation:We are given:$\operatorname{cosec}\theta - \sin\theta = m$ and $\sec\theta - \cos\theta = n$.Rewrite each in simpler forms:$\operatorname{cosec}\theta - \sin\theta = \frac{1}{\sin\theta} - \sin\theta = \frac{1 - \sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta}$.Thus $m = \frac{\cos^2\theta}{\sin\theta}$.Similarly, $\sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$.Thus $n = \frac{\sin^2\theta}{\cos\theta}$.We need the value of $m^{4/3}n^{2/3} + m^{2/3}n^{4/3}$.Factor the expression: $m^{4/3}n^{2/3} + m^{2/3}n^{4/3} = m^{2/3}n^{2/3}(m^{2/3} + n^{2/3})$.Now, substitute $m = \frac{\cos^2\theta}{\sin\theta}$ and $n = \frac{\sin^2\theta}{\cos\theta}$:$m^{2/3}n^{2/3} = \left(\frac{\cos^2\theta}{\sin\theta}\right)^{2/3} \cdot \left(\frac{\sin^2\theta}{\cos\theta}\right)^{2/3} = \frac{\cos^{4/3}\theta}{\sin^{2/3}\theta} \cdot \frac{\sin^{4/3}\theta}{\cos^{2/3}\theta} = \cos^{(4/3 - 2/3)}\theta \cdot \sin^{(4/3 - 2/3)}\theta = \cos^{2/3}\theta \cdot \sin^{2/3}\theta$.Also, $m^{2/3} + n^{2/3} = \left(\frac{\cos^2\theta}{\sin\theta}\right)^{2/3} + \left(\frac{\sin^2\theta}{\cos\theta}\right)^{2/3} = \frac{\cos^{4/3}\theta}{\sin^{2/3}\theta} + \frac{\sin^{4/3}\theta}{\cos^{2/3}\theta}$.Multiply $m^{2/3}n^{2/3}$ with $(m^{2/3}+n^{2/3})$:$m^{4/3}n^{2/3} + m^{2/3}n^{4/3} = \cos^{2/3}\theta \sin^{2/3}\theta \left( \frac{\cos^{4/3}\theta}{\sin^{2/3}\theta} + \frac{\sin^{4/3}\theta}{\cos^{2/3}\theta} \right)$.Simplify: $= \cos^{2/3}\theta \sin^{2/3}\theta \cdot \frac{\cos^{4/3}\theta}{\sin^{2/3}\theta} + \cos^{2/3}\theta \sin^{2/3}\theta \cdot \frac{\sin^{4/3}\theta}{\cos^{2/3}\theta}$$= \cos^{2/3+4/3}\theta \cdot \sin^{2/3-2/3}\theta + \cos^{2/3-2/3}\theta \cdot \sin^{2/3+4/3}\theta$$= \cos^{2}\theta \cdot \sin^{0}\theta + \cos^{0}\theta \cdot \sin^{2}\theta$$= \cos^{2}\theta + \sin^{2}\theta = 1$.Thus the expression equals $1$ for all $\theta$ where defined.Answer:$1$ (Option B).