1.

`a(cosC-cosB)=2(b-c)cos^2A/2`

Answer» From sine law, we have,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
Now, `L.H.S. = a(cosC-cosB) = ksinA(cosC-cosB)`
`=k(2sin(A/2)cos(A/2))[2sin((B+C)/2)sin((B-C)/2)]`
`=k(2sin(A/2)cos(A/2))[2sin((pi-A)/2)sin((B-C)/2)]...[As A+B+C = pi]`
`=4ksin((pi-(B+C))/2)cos(A/2)cos(A/2)sin((B-C)/2)`
`=4kcos^2(A/2)cos((B+C)/2)sin((B-C)/2)`
`=2kcos^2(A/2)(2cos((B+C)/2)sin((B-C)/2))`
`=2kcos^2(A/2)(sinB-sinC)`
`=2kcos^2(A/2)(b/k-c/k)...[b = ksinB, c = ksinC]`
`=2cos^(A/2)(b-c) = R.H.S.`


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