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If A,B and C are angles of a triangle ,then the determinant$\begin{vmatrix}1 & \cos C & \cos B\\\cos C & 1 & \cos A\\\cos B & \cos A & 1\end{vmatrix}$ is equal to

$\begin{array}{1 1}(A)\quad 0 & (B)\quad -1\\(C)\quad 1 & (D)\quad None\;of\;these\end{array}$

1 Answer

  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
Let $\Delta=\begin{vmatrix}1 & cos C & cos B\\cos C & 1 &cos A\\cos B& cos A & 1\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=1(1-cos^2A)-cos C(cos C-cos A cos B)+cos B(cos Ccos A-cos B)$
$\quad=1-cos^2A-cos^2C+cos Acos B cos C+cos A cos B cos C-cos^2A$
$\quad=1+2cos Acos Bcos C-(cos^2A+cos^2B+cos ^2C)$
But $cos ^2A+cos^2B+cos^2C=1+2cos A cosB cos C$
Substituting this we get,
$\Delta=1+2cos A cos B cos C-1-2cos A cos B cos C=0.$
Hence $\Delta=0.$
answered Mar 26, 2013 by sreemathi.v

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