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Prove that a sinA - b sinB = c sin(A-B) |
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Answer» Solution :R.H.S = c SIN (A-B) = 2R sinC sin (A-B) 2R sin (A+B) sin (A-B) `[THEREFORE A + B + C = pi` or `A + B = pi - C` or sin (A + B) = sin (pi-C) sinC] = 2R `(sin^2A - sin^2B)` = 2R sinA sinA - 2R SINB sinB a sinA - b sinB = L.H.S |
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