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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
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B
1
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C
-1
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D
2
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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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