1.

Prove that a sinA - b sinB = c sin(A-B)

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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