Math, asked by s11948, 9 months ago

Prove that: sin(A−C)+2 sin A+sin(A+C)/sin(B−C)+2 sin B+sin(B+C)=sin A/sin B

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Answered by Anonymous

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[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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Answered by tennetiraj86

Answer:

answer for the given problem is given

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