1.

For any triangle ABC, prove that`(a-b)/c=(sin((A-B)/2))/(cosC/2)`

Answer» From sine law, we have,`a/sinA = b/sinB = c/sinC = k`
So, `a = ksinA, b = ksinB, c = ksinC`
`:. L.H.S. = (a-b)/c = (k(sinA - sinB))/(ksinC)`
`= (sinA- sinB)/(sinC)`
`=(2sin((A-B)/2)cos((A+B)/2))/(2sin(C/2)cos(C/2))`
Here, `A+B+C = 180. So, A+B = 180- C`
So, our expression becomes,
`= (sin((A-B)/2)cos((180-C)/2))/(sin(C/2)cos(C/2))`
`= (sin((A-B)/2)cos(90-C/2))/(sin(C/2)cos(C/2))`
`= (sin((A-B)/2)sin(C/2))/(sin(C/2)cos(C/2))`
`= sin((A-B)/2)/cos(C/2) = R.H.S.`


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