If A, B, C be the angles of a triangle, | vedclass

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Question

(b) Given, Angles of a triangle $= A, B$ and $C$. We know that as $A + B + C = \pi$,

therefore $A + B = \pi - C$

or $\cos (A + B) = \cos (\pi -

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(b) Given, Angles of a triangle $= A, B$ and $C$. We know that as $A + B + C = \pi$,

therefore $A + B = \pi - C$

or $\cos (A + B) = \cos (\pi - C) = - \cos C$

or $\cos A\cos B - \sin A\sin B = - \cos C$

$\cos A\cos B + \cos C = \sin A\sin B$

and  $\sin (A + B) = \sin (\pi - C) = \sin C.$

Expanding the given determinant, we get

$\Delta = - (1 - {\cos ^2}A) + \cos C(\cos C + \cos A\cos B)$

$ + \cos B(\cos B + \cos A\cos C)$

$ = - {\sin ^2}A + \cos C(\sin A\sin B) + \cos B(\sin A\sin C)$

$ = - {\sin ^2}A + \sin A(\sin B\cos C + \cos B\sin C)$

$ = - {\sin ^2}A + \sin A\sin (B + C)$$ = - {\sin ^2}A + {\sin ^2}A = 0.$

If $A, B, C$ be the angles of a triangle, then $\left| {\,\begin{array}{*{20}{c}}{ - 1}&{\cos C}&{\cos B}\\{\cos C}&{ - 1}&{\cos A}\\{\cos B}&{\cos A}&{ - 1}\end{array}\,} \right| = $

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