(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.$