If A+B+C=180° , prove that sin²A+sin²B+sin²C=2(1+cosAcosBcosC)
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Answered by
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I'm attaching a picture of the answer
Hope it helps...and don't forget to mark the answer brainliest if you really find it useful.
doubts are always welcome...
Thanks
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neerajvermag11:
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Cos²A-sin²B =cos(A+B)cos(A-B) how
Please tell me Bhai
Find derivative of tan inverse (cosx/1+sinx) w.r.t.x
Bhai please solve this
Answered by
10
Given,
A+B+C=180° , prove sin²A+sin²B+sin²C=2(1+cosAcosBcosC)
Let's LHS = sin2A + sin2B + sin2C
so that
2S = 2sin2A + 1 – cos2B +1 – cos2C
= 2 sin2A + 2 – 2cos(B + C)
cos(B – C)
= 2 – 2 cos2A + 2
– 2cos(B + C) cos(B – C)
∴ S = 2 + cosA [cos(B – C) + cos(B+
C)] since cosA
= – cos(B+C)
∴ S = 2 + 2 cos A cos B cos C
= 2(1+ cos A cos B cos C)
= RHS
Therefore, LHS = RHS is
proved.
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