If A+B+C=3π2,then cos2A+cos⁡2B+cos⁡2C is equal to

A
$1 - 4 \cos Acos Bcos C$
B
$4 \sin Asin Bsin C$
C
$1 + 2 \cos Acos Bcos C$
D
$1 - 4 \sin Asin Bsin C$

Solution:

Since, $A + B + C = \frac{3 π}{2}$

$\begin{array}{l} ∴ \cos 2 A + \cos 2 B + \cos 2 C \\ = 2 \cos (A + B) \cos (A - B) + \cos 2 C \\ = 2 \cos (\frac{3 π}{2} - C) \cos (A - B) + 1 - 2 \sin^{2} C \\ = 1 - 2 \sin C [\cos (A - B) + \sin (\frac{3 π}{2} - (A + B))] \\ = 1 - 2 \sin C [\cos (A - B) - \cos (A + B)] \\ = 1 - 4 \sin Asin Bsin C \end{array}$

If $A + B + C = \frac{3 π}{2}$ ,then cos2 $A + \cos 2 B + \cos 2 C$ is equal to

Solution:

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