If A + B + C = {270^o}, then cos ,2A + cos 2B + cos 2C + 4sin A,sin B,sin C =

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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

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