A, B, C are the angles of a triangle, then {s | vedclass

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Question

(b) ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C$ 

$ = 1 - {\cos ^2}A + 1 - {\cos ^2}B + {\sin ^2}C$

$ = 2 - {\cos ^2}A - \cos (B + C)\cos (B - C)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b) ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C$ 

$ = 1 - {\cos ^2}A + 1 - {\cos ^2}B + {\sin ^2}C$

$ = 2 - {\cos ^2}A - \cos (B + C)\cos (B - C)$

$ = 2 - \cos A[\cos A - \cos (B - C)]$

$ = 2 - \cos A[ - \cos (B + C) - \cos (B - C)]$ 

$ = 2 + \cos A.2\cos B\cos C$  

$	herefore$ ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\cos B\cos C = 2$.

$A, B, C$ are the angles of a triangle, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\,\cos B\,\cos C = $

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