#### Question

In ΔABC, prove that : `tan((a-b)/2)=(a-b)/(a+b)cotC/2`

#### Solution

In ΔABC by sine rule, we have

`a/sinA=b/sinB=c/sinC=k`

a=ksinA,b=ksinB and c=ksinC

Now, consider

`(a-b)/(a+b)=(ksinA-ksinB)/(ksinA+ksinB)`

`=(sinA-sinB)/(sinA+sinB)`

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

`=cot((A+B)/2).tan((A-B)/2)`

`=cot(pi/2-C/2).tan((A-B)/2) .....[because A+B+C=pi]`

`=tan(C/2)tan((A-B)/2)`

`(a-b)/(a+b)=tan((A-B)/2)/cot(C/2)`

`tan((A-B)/2)=(a-b)/(a+b)cot(C/2)`

Is there an error in this question or solution?

#### APPEARS IN

Solution In ΔABC, prove that : tan((a-b)/2)=(a-b)/(a+b) cotC/2 Concept: Trigonometric Functions - Solution of a Triangle.