Question

is anyone able to solve this????????​

Answer 1

Consider the problem,

sin2A+sin2B+sin2c=2+2cosA.cosB.cosC

We can write sin2A as,

sin2A=21−cos(2A)

Therefore,

LHS=21−cos(2A)+21−cos(2B)+21−cos(2C)=23−(cos(2A)+cos(2B)+cos(2C))=21(3−(2cos(A+B)cos(A−B)+cos(2C)))C=180−(A+B)cos(C)=cos(180−(A+B))cos(C)=−cos(A+B)

Therefore,

=23−(−2cos(C)cos(A−B)+cos(2C))cos(2C)=2

Answer 2

Step-by-step explanation:

This is your answer... check the photos...

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