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sin(a-c) + 2sinA + SIN(A+C) / SIN(B-C) + 2sin B + Sin (b+c) = sinA/sinB

Rahul kumar , 7 Years ago
Grade 11
anser 1 Answers
Akash Yadav

Last Activity: 6 Years ago

[sin(A-C)+2sinA+sin(A+C)]/[sin(B-C)+2sinB+sin(B+C)].
=[2sinA+sin(A+C)+sin(A-C)]/[2sinB+sin(B+C)+sin(B-C)].
=[2sinA+2sin(A+C+A-C)/2.cos(A+C-A+C)/2]/[2.sinB+2sin(B+C+B-C)/2.cos(B+C-B+C)/2].
=(2sinA+2sinA.cosC)/(2sinB+2sinB.cosC)
=2sinA(1+cosC)/2sinB(1+cosC)
=sinA/sinB , Answer.

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