In `DeltaABC`, prove that:
`(b-c)/(a)=(tanB/2-tanC/2)/(tanB/2+tanC/2)`

Explanation

LHS `=(b-c)/a=(ksinB-ksinC)/(ksinA)`
`=(sinB-sinC)/(sinA)=(2cos(B+C)/(2).sin(B-C)/(2))/(2sinA/2.cosA/2)`
`=(cos(B+C)/(2).sin(B-C)/(2))/(sin(180^(@)-(B+C))/(2).cos(180^(@)-(B+C))/(2))`
`(cos(B+C)/(2).sin(B-C)/(2))/(sin{90^(@)-(B+C)/(2)}.cos{90^(@)-(B+C)/(2)}`
`=(cos(B+C)/(2).sin(B-C)/(2))/(cos(B+C)/(2).sin(B+C)/(2))=(sin(B/2-C/2)/(sin(B/2+C/2)`
`=sinB/2cosC/2-cosB/2sinC/2)/(sinB/2cosC/2+cosB/2sinC/2)`
`=(tanB/2-tanC/2)/(tanB/2+tanC/2)` (Divide `N^(r)` and `D^(r)` by `cosB/2cosC/2`)
=RHS Hence Proved.