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Prove cot A/2 + cot B/2 + cot C/2 = S^2/Δ
(a+b)/2=(pi-c)/2cot[(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/2as c=cot and t=tan substitutingcot 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
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