Prove cot A/2 + cot B/2 + cot C/2 = S^2/Δ

(a+b)/2=(pi-c)/2

cot[(a+b)/2]=cot(90-c/2)
To save space I will write cot as 'c' and tan as 't'
[c(a/2)c(b/2)-1]/[c(a/2)+c(b/2)]=t(c/2…

cross multiplying

[c(a/2)c(b/2)-1]*(1/t(c/2)=c(b/2)+c(b/…

[c(a/2)c(b/2)-1]*c(c/2)=c(b/2)+c(b/2)

c(a/2)c(b/2)c(c/2)-c(c/2)=c(b/2)+c(b/2…

c(a/2)c(b/2)c(c/2)=c A/2.c B/2.c C/2

as c=cot and t=tan substituting

cot A/2+cot B/2+ cot C/2=cot A/2.cot B/2.cot C/2

Now substitute the formulae of cot A/2,cot B/2 and cot C/2 and you can get the required answer

Thanx