In triangle `A B C ,(sinA+sinB+sinC)/(sinA+sinB-sinC)` is equal to

Explanation

`A+B+C=theta`

`sin((A+B)/2)=cos(C/2)`

`D->sinA+sinB-sinC`

`=2sin((A+B)/2)cos((A-B)/2)-sinC`

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

`=2cos(C/2)[cos((A-B)/2)-sin(C/2)]`

`=2cos(C/2)[cos((A-B)/2)-cos((A+B)/2)]`

`=2cos(C/2)[2sin(A/2)sin(B/2)]`

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