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3.Trigonometrical Ratios, Functions and Identities
easy
If $A + B + C = {270^o},$ then $\cos \,2A + \cos 2B + \cos 2C + 4\sin A\,\sin B\,\sin C = $
A
$0$
B
$1$
C
$2$
D
$3$
Solution
(b) $A + B + C = {270^o}\,\,\, \Rightarrow A = B = C = {90^o},$ then
$\cos 2A + \cos 2B + \cos 2C + 4\sin A\,\sin B\,\sin C$
$ = \cos {180^o} + \cos {180^o} + \cos {180^o} + 4\sin {90^o}\sin {90^o}\sin {90^o}$
$ = – 1 – 1 – 1 + 4 \cdot 1\cdot 1\cdot 1 = – 3 + 4 = 1$.
Standard 11
Mathematics