UPSC CDS I 2025 Mathematics Paper

Concept:Use the given equation to express $\sin\theta$ in terms of $m$ and $n$, then substitute into the target trigonometric expression and simplify.Explanation:Start with the given equation: $m^{2}(\sin\theta-1)+n^{2}(\sin\theta+1)=0$.Expand and simplify: $m^{2}\sin\theta-m^{2}+n^{2}\sin\theta+n^{2}=0$.Combine like terms: $(m^{2}+n^{2})\sin\theta + (n^{2}-m^{2})=0$.So $(m^{2}+n^{2})\sin\theta = m^{2}-n^{2}$.Thus $\sin\theta = \dfrac{m^{2}-n^{2}}{m^{2}+n^{2}}$.Now the target expression: $(m^{2}+n^{2})\cos\theta - (m^{2}-n^{2})\cot\theta$.Recall $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$.Substitute $\sin\theta$: $(m^{2}+n^{2})\cos\theta - (m^{2}-n^{2})\times \dfrac{\cos\theta}{\dfrac{m^{2}-n^{2}}{m^{2}+n^{2}}}$.Simplify the second term: $(m^{2}-n^{2})\cos\theta \times \dfrac{m^{2}+n^{2}}{m^{2}-n^{2}} = (m^{2}+n^{2})\cos\theta$.Therefore the expression becomes $(m^{2}+n^{2})\cos\theta - (m^{2}+n^{2})\cos\theta = 0$.No further simplification needed; the value is independent of $\theta$.Answer:0 (Option D)