UPSC NDA II 2018 Mathematics Solved Paper

Concept:A determinant becomes zero when any two rows or columns are identical; applying elementary row/column operations does not change its value.Explanation:We start with the given determinant equal to zero:$$\begin{vmatrix} 1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A+\sin^{2}A & \sin B+\sin^{2}B & \sin C+\sin^{2}C \end{vmatrix}=0$$Perform row operations: $R_{1}\rightarrow R_{1}-R_{2}$ and $R_{3}\rightarrow R_{3}-R_{2}$:$$\begin{vmatrix} -\sin A & -\sin B & -\sin C \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin^{2}A-1 & \sin^{2}B-1 & \sin^{2}C-1 \end{vmatrix}=0$$Replace $\sin^{2}A-1$ with $-\cos^{2}A$ (using $\sin^{2}\theta+\cos^{2}\theta=1$):$$\begin{vmatrix} -\sin A & -\sin B & -\sin C \\ 1+\sin A & 1+\sin B & 1+\sin C \\ -\cos^{2}A & -\cos^{2}B & -\cos^{2}C \end{vmatrix}=0$$Apply $R_{2}\rightarrow R_{2}+R_{1}$:$$\begin{vmatrix} -\sin A & -\sin B & -\sin C \\ 1 & 1 & 1 \\ -\cos^{2}A & -\cos^{2}B & -\cos^{2}C \end{vmatrix}=0$$Apply $R_{3}\rightarrow R_{3}+R_{2}$:$$\begin{vmatrix} -\sin A & -\sin B & -\sin C \\ 1 & 1 & 1 \\ -\cos^{2}A+1 & -\cos^{2}B+1 & -\cos^{2}C+1 \end{vmatrix}=0$$Since $-\cos^{2}A+1=\sin^{2}A$, we get:$$\begin{vmatrix} -\sin A & -\sin B & -\sin C \\ 1 & 1 & 1 \\ \sin^{2}A & \sin^{2}B & \sin^{2}C \end{vmatrix}=0$$Now apply column operations: $C_{2}\rightarrow C_{2}-C_{1}$ and $C_{3}\rightarrow C_{3}-C_{2}$:$$\begin{vmatrix} -\sin A & \sin A-\sin B & \sin B-\sin C \\ 1 & 0 & 0 \\ \sin^{2}A & \sin^{2}B-\sin^{2}A & \sin^{2}C-\sin^{2}B \end{vmatrix}=0$$Expand along the second row (which has only one non‑zero entry):$$-1\cdot\Big[(\sin A-\sin B)(\sin^{2}C-\sin^{2}B)-(\sin B-\sin C)(\sin^{2}B-\sin^{2}A)\Big]=0$$Factor differences of squares: $\sin^{2}C-\sin^{2}B=(\sin C-\sin B)(\sin C+\sin B)$ and similarly for the other term:$$-(\sin A-\sin B)(\sin C-\sin B)(\sin C+\sin B) + (\sin B-\sin C)(\sin A-\sin B)(\sin A+\sin B)=0$$Factor out $(\sin A-\sin B)(\sin C-\sin B)$:$$(\sin A-\sin B)(\sin C-\sin B)\big[-(\sin C+\sin B)+(\sin A+\sin B)\big]=0$$Simplify the bracket: $-\sin C-\sin B+\sin A+\sin B = \sin A-\sin C$. Thus:$$(\sin A-\sin B)(\sin C-\sin B)(\sin A-\sin C)=0$$Hence at least one factor is zero: $\sin A=\sin B$, $\sin C=\sin B$, or $\sin A=\sin C$. Since $A,B,C$ are angles of a triangle (each between $0$ and $\pi$), sine equality implies angle equality. Therefore $A=B=C$.Answer:The triangle ABC is equilateral, so option B is correct.