Question

In triangle ABC, prove that the area of the in-circle is to the area of the triangle itself $=\pi :\cot \left(A/2\right)\cot \left(B/2\right)\cot \left(C/2\right).$

Answer 1

$\frac{areaofcircle}{areaoftriangle}=\frac{\pi r^2}{S}=\frac{\pi }{S}.\frac{S^2}{s^2}=\pi .\frac{S_2}{s^2}$
Not cot (A/2). cot (B/2). cot (C/2)
$={[\frac{s(s-a)}{(s-b)(s-c)}.\frac{s(s-b)}{(s-c)(s-a)}.\frac{s(s-c)}{(s-a)(s-b)}]}^{1/2}$
$={\left[\frac{s^3}{(s-a)}(s-b)(s-c)\right]}^{1/2}$
$={\left[\frac{s^4}{s(s-a)}(s-b)(s-c)\right]}^{1/2}=\frac{s^2}{S}.$
Now
$\frac{\pi }{cot\left(A/2\right)cot\left(B/2\right)cot\left(C/2\right)}=\frac{\pi }{s^2/S}=\frac{\pi S}{s^2}.$
Hence from (1) and (2), we prove the required relation.