State true or false.
In any triangle $A B C$ , $s i n 2 A + s i n 2 B + s i n 2 C = - 4 s i n A s i n B s i n C$

True

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False

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Solution

The correct option is B False
$sin 2 A + sin 2 B + sin 2 C$

$= (sin 2 A + sin 2 B) + sin 2 C$

$= 2 sin (\frac{2 A + 2 B}{2}) cos (\frac{2 A - 2 B}{2}) + 2 sin C cos C$

$= 2 sin (A + B) cos (A - B) + 2 sin C cos C$

$= 2 sin (π - C) cos (A - B) + 2 sin C cos C$ since

$A + B + C = π$

$= 2 sin C cos (A - B) + 2 sin C cos (π - (A + B))$ since $A + B + C = π$

$= 2 sin C [cos (A - B) - cos (A + B)]$

$= 2 sin C [- 2 sin (\frac{A - B + A + B}{2}) sin (\frac{A - B - A - B}{2})]$

$= - 4 sin C [sin A sin (- B)]$

$= 4 sin A sin B sin C$

Hence the given statement is false.

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