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For any triangle ABC, prove that`(b+c)cos((B+C)/2)=acos((B-C)/2)`

Answer» We know,`a/sinA = b/sinB = c/sinC = k`
where, k is a constant.
`:. (b+c)/a = (k(sinB+sinC))/(ksinA)`
`= (sinB+sinC)/sinA`
`=(2sin((B+C)/2)cos((B-C)/2))/(2sinA/2cosA/2)`
As, `A+B+C = 180, So, B+C = 180-A`
`:. (b+c)/a=(sin((180-A)/2)cos((B-C)/2))/(sin((180-(B+C))/2)cosA/2)`
`=(sin(90-A/2)cos((B-C)/2))/(sin(90-(B+C)/2)cosA/2)`
`=(cos(A/2)cos((B-C)/2))/(cos((B+C)/2)cosA/2)`
`(b+c)/a= (cos((B-C)/2))/(cos((B+C)/2))`
So, `(b+c)(cos((B+C)/2)) = a(cos((B-C)/2))`


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