Question:
If \( \alpha,\beta,\gamma \) are the angles of a triangle, then the value of the determinant \( \begin{vmatrix}\cot\alpha&\cot\beta&\cot\gamma\\1&1&1\\\sin^{2}\alpha&\sin^{2}\beta&\sin^{2}\gamma\end{vmatrix} \) is equal to ________
If \( \alpha,\beta,\gamma \) are the angles of a triangle, then the value of the determinant \( \begin{vmatrix}\cot\alpha&\cot\beta&\cot\gamma\\1&1&1\\\sin^{2}\alpha&\sin^{2}\beta&\sin^{2}\gamma\end{vmatrix} \) is equal to ________
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Expand along the first row, use \( \sin^2\gamma-\sin^2\beta=\sin(\gamma-\beta)\sin\alpha \) and \( \alpha+\beta+\gamma=\pi \).
Updated On: Nov 12, 2026
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Solution and Explanation
Expanding along the first row: \( D=\cot\alpha(\sin^2\gamma-\sin^2\beta)-\cot\beta(\sin^2\gamma-\sin^2\alpha)+\cot\gamma(\sin^2\beta-\sin^2\alpha) \).
Use \( \sin^2\gamma-\sin^2\beta=\sin(\gamma+\beta)\sin(\gamma-\beta)=\sin\alpha\sin(\gamma-\beta) \) etc. Then \( \cot\alpha(\sin^2\gamma-\sin^2\beta)=\cos\alpha\sin(\gamma-\beta) \).
So \( D=\cos\alpha\sin(\gamma-\beta)-\cos\beta\sin(\gamma-\alpha)+\cos\gamma\sin(\beta-\alpha) \).
With \( \cos\alpha=-\cos(\beta+\gamma) \) etc., the three terms become \( -\frac12(\sin2\gamma-\sin2\beta) \), \( +\frac12(\sin2\gamma-\sin2\alpha) \) and \( -\frac12(\sin2\beta-\sin2\alpha) \), which sum to 0.
Answer: 0
Use \( \sin^2\gamma-\sin^2\beta=\sin(\gamma+\beta)\sin(\gamma-\beta)=\sin\alpha\sin(\gamma-\beta) \) etc. Then \( \cot\alpha(\sin^2\gamma-\sin^2\beta)=\cos\alpha\sin(\gamma-\beta) \).
So \( D=\cos\alpha\sin(\gamma-\beta)-\cos\beta\sin(\gamma-\alpha)+\cos\gamma\sin(\beta-\alpha) \).
With \( \cos\alpha=-\cos(\beta+\gamma) \) etc., the three terms become \( -\frac12(\sin2\gamma-\sin2\beta) \), \( +\frac12(\sin2\gamma-\sin2\alpha) \) and \( -\frac12(\sin2\beta-\sin2\alpha) \), which sum to 0.
Answer: 0
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