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February 14th, 2018, 09:52 AM   #1
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Prove: (Sin B + sin C - sin A)/(sin A + sin C - sin B) = tan(B/2) . cot (A/2)

Hello everyone!

I am preparing for a test and I got to this question in my coursebook. However, I'm not able to solve it. Can someone help me?

I already tried using the sine law on the left side, the definition of tangens and cotangens on the right side. Nothing got me to prove it.

Can someone help me please?

Thank you in advance!
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February 14th, 2018, 11:45 AM   #2
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You need the additional information that sin C = sin(A + B). Perhaps the book tells you that A, B and C are the angles of a triangle for this problem or for a group of questions that includes this problem.

(sin B + sin C - sin A)/(sin A + sin C - sin B)
= (sin B + sin(A + B) - sin A)/(sin A + sin(A + B) - sin B)
= (2sin(B/2)cos(B/2) + 2cos(A + B/2)sin(B/2))/(2sin(A + B/2)cos(B/2) - 2sin(B/2)cos(B/2))
= tan(B/2)(cos(B/2) + cos(A + B/2))/(sin(A + B/2) - sin(B/2))
= tan(B/2)(2cos((A + B)/2)cos(A/2))/(2cos((A + B)/2)sin(A/2))
= tan(B/2)cot(A/2)
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