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3.Trigonometrical Ratios, Functions and Identities
medium
$A, B, C$ are the angles of a triangle, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\,\cos B\,\cos C = $
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(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$
$\therefore$ ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C – 2\cos A\cos B\cos C = 2$.
Standard 11
Mathematics
