For any triangle ABC, prove that`(a+b)/c=(cos((A-B)/2))/(sinC/2)`

Explanation

By sine rule,
`a/(sinA) = b/(sinB)= c/(sinC)=k` (say)
`therefore a = k sinA, b=k sinB` and `c= ksinC`
Now, LHS `=(a+b)/( c) = (ksinA + k sinB)/( k sinC)`
`=(sinA+sinB)/(sin C)`
`=(2sin(A+B)/2 cos(A-B)/(2))/(2sinC/2cosC/2)`
`=(cosC/2 cos(A-B)/(2))/(sinC/2cosC/2)`
`=(cos(A-B)/(2))/(sinC/2)= RHS`
Hence proved.