Question

In a triangle $ABC,$ if$\cot \left(\frac{A}{2}\right)\cot \left(\frac{B}{2}\right)=c,$$\cot \left(\frac{B}{2}\right)\cot \left(\frac{C}{2}\right)=a,$
and $\cot \left(\frac{C}{2}\right)\cot \left(\frac{A}{2}\right)=b$, then$\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}=$

Answer 1

Given In a triangle $ABC,$
$\cot \left(\frac{A}{2}\right)\cot \left(\frac{B}{2}\right)=c$
$\Rightarrow \sqrt{\frac{s(s-a)}{(s-b)(s-c)}\times \frac{s(s-b)}{(s-c)(s-a)}}=c$
$\Rightarrow \frac{s}{s-c}=c$
$\Rightarrow \frac{1}{s-c}=\frac{c}{s}$
Again $\cot \left(\frac{B}{2}\right)\cot \left(\frac{C}{2}\right)=a$
$\Rightarrow \sqrt{\frac{s(s-b)}{(s-c)(s-a)}\times \frac{s(s-c)}{(s-a)(s-b)}}=a$
$\Rightarrow \frac{s}{s-a}=a$
$\Rightarrow \frac{1}{s-a}=\frac{a}{s}$
and $\cot \left(\frac{C}{2}\right)\cot \left(\frac{A}{2}\right)=b$
$\Rightarrow \sqrt{\frac{s(s-c)}{(s-a)(s-b)}\times \frac{s(s-a)}{(s-b)(s-c)}}=b$
$\Rightarrow \frac{s}{s-b}=b$
$\Rightarrow \frac{1}{s-b}=\frac{b}{s}$
So that $\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}$
$=\frac{a}{s}+\frac{b}{s}+\frac{c}{s}$
$=\frac{a+b+c}{s}=\frac{2s}{s}=2$