Question

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Answer 1

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If A+B+C = π, then P.T. sin2A + sin2B + in2C = 4 sinAsinBsinC.

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Brother your correct answer will be☟

IN STEP-BY-STEP-EXPLANATION:-

$\begin{gathered} \sin(2a) + \sin(2b) + \sin(2c) \\ 2\sin( \frac{2a + 2b}{2} ) \cos( \frac{2a - 2b}{2} ) + \sin(2c) \\ 2 \sin(a + b) \cos(a - b) + \sin(2c) \\ a + b + c = 180 \\ a + b = 180 - c \\ 2 \sin(180 - c) \cos(a - b) + \sin(2c) \\ 2 \sin(c) \cos(a - b) + 2 \sin(c) \cos(c) \\ 2 \sin(c) ( \cos(a - b) + \cos(c) ) \\ 2 \sin(c) \times 2 \cos( \frac{a - b + c}{2} ) \cos( \frac{a - b - c}{2} ) \\ 4 \sin(c) \times \cos( \frac{a +c - b}{2} ) \times \cos( \frac{a - (b + c)}{2} ) \\ 4 \sin(c) \times \cos(90 - b ) \times \cos(90 - a) \\ 4 \sin(c) \times \sin(b) \times \sin(a)\end{gathered}$

So Now it is Proved.

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✌︎Thanking you, your one of the Brother of your 130 Crore Indian Brothers and Sisters.✌︎

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Answer 2

Answer:

sin(

2

2a+2b

)cos(

2

2a−2b

)+sin(2c)

2sin(a+b)cos(a−b)+sin(2c)

a+b+c=180

a+b=180−c

2sin(180−c)cos(a−b)+sin(2c)

2sin(c)cos(a−b)+2sin(c)cos(c)

2sin(c)(cos(a−b)+cos(c))

2sin(c)×2cos(

2

a−b+c

)cos(

2

a−b−c

)

4sin(c)×cos(

2

a+c−b

)×cos(

2

a−(b+c)

)

4sin(c)×cos(90−b)×cos(90−a)

4sin(c)×sin(b)×sin(a)