Question

# lf $$A+B+C=\pi$$, then$$\begin{vmatrix}\sin^{2}A & \cot A & 1\\ \sin^{2}B & \cot B & 1\\ \sin^{2}C & \cot C & 1\end{vmatrix}$$

A
0
B
1
C
-1
D
2

Solution

## The correct option is A 0$$\begin{vmatrix}\sin^{2}A & \cot A & 1\\ \sin^{2}B & \cot B & 1\\ \sin^{2}C & \cot C & 1\end{vmatrix}$$$${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2, }{ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 }$$$$=\begin{vmatrix} \sin ^{ 2 } A-\sin ^{ 2 } B & \cot A-\cot { B } & 0 \\ \sin ^{ 2 } B-\sin ^{ 2 } C & \cot B-\cot { C } & 0 \\ \sin ^{ 2 } C & \cot C & 1 \end{vmatrix}$$$$=\begin{vmatrix} \sin { (A+B) } \sin { (A-B) } & \dfrac { \cos A\sin { B } -\sin { A\cos { B } } }{ \sin { A } \sin { B } } & 0 \\ \sin { (B+C) } \sin { (B-C) } & \dfrac { \cos B\sin { C } -\sin { B\cos { C } } }{ \sin { B } \sin { C } } & 0 \\ \sin ^{ 2 } C & \cot C & 1 \end{vmatrix}$$$$=\begin{vmatrix} \sin { C } \sin { (A-B) } & -\dfrac { \sin { (A-B) } }{ \sin { A } \sin { B } } & 0 \\ \sin { A } \sin { (B-C) } & -\dfrac { \sin { (B-C) } }{ \sin { B } \sin { C } } & 0 \\ \sin ^{ 2 } C & \cot C & 1 \end{vmatrix}$$$$\quad =-\dfrac { \sin { (B-C)\sin { (A-B) } } }{ \sin { B } } +\dfrac { \sin { (B-C)\sin { (A-B) } } }{ \sin { B } }$$$$=0$$Mathematics

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