In any `triangle ABC`, prove that than `((A-B)/(2))=((a-b)/(a+b))cot .(C)/(2)`.

Explanation

We know that, `(a)/(sinA)=(b)/(sinB)=K`
`a=KsinA,b=KsinB`
`(a-b)/(a+b)=(KsinA-Ksinb)/(KsinA+KsinB)`
`=(sinA-sinB)/(sinA+sinB)`
`=(2"cos"(A+B)/(2)"sin"(A-B)/(2))/(2"sin"(A+B)/(2)"cos"(A-B)/(2))`
`(a-b)/(a+b)=("tan"(A-B)/(2))/("tan"(A+B)/(2))`
`="cot"(A+B)/(2)"tan"(A-B)/(2)`
We know that,
`(A+B+C)/(2)=90^(@),(A+B)/(2)=90^(@)-(C)/(2)`
`cot((A+B)/(2))=cot(90^(@)-(C)/(2))`
`="tan"(C)/(2)`
`(a-b)/(a+b)="tan"(C)/(2)"tan"(A-B)/(2)`
`"tan"(A-B)/(2)=((a-b)/(a+b))"cot"(C)/(2)`