1.

`sin ((C-A)/2) =(c-a)/b cos(B/2)`

Answer» `a/sinA=b/sinB=c/sinA=k`
`a=ksinA`
`b=ksinB`
`c=ksinC`
RHS
`=(ksinC-ksinA)/(ksinB)*cos(B/2)`
`=(sinC-sinA)/sinB*cos(B/2)`
`=(sin((C-A)/2)cos((C+A)/2))/(sin(B/2)cos(B/2))*cos(B/2)`
`=sin((C-A)/2)[cos((C+A)/2)]/sin(B/2)`
`=sin((C-A)/2)`
LHS.


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