1.

For any triangle ABC, prove that`(a+b)/c=(cos((A-B)/2))/(sinC/2)`

Answer» let `a/sinA = b/sin B = c/Sin C = k` (say)
`(a+b)/c = (ksinA + ksinB)/(ksinC) `
`= (sinA + sinB)/(sinC)`
`= (2sin((A+B)/2)cos((A-B)/2))/(2sin(C/2)cos(C/2))`
``
as A+B+C = `180^@`
`= (sin(90^@ - C/2)cos((A-B)/2))/(sin(C/2)cos(C/2))`
`= (cos(C/2).cos((A-B)/2))/(sin(C/2)cos(C/2)) `
`= (Cos((A-B)/2))/Sin(c/2)`


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