Question

Prove that

$\frac{(a+b+c{)}^2}{a^2+b^2+c^2}=\frac{cot\frac{1}{2}A+cot\frac{1}{2}B+cot\frac{1}{2}C}{cotA+cotB+cotC}$

Answer 1

$\cot \frac{1}{2}A+cot\frac{1}{2}B+cot\frac{1}{2}C$

= $\sqrt{[\frac{s(s-a)}{(s-a)(s-c)}]}+\sqrt{[\frac{s(s-b)}{(s-c)(s-a)}]}+\sqrt{[\frac{s(s-c)}{(s-a)(s-b)}]}$

= $\frac{\sqrt{s}\left[\right(s-a)+(s-b)+(s-c\left)\right]}{\sqrt{s}\left[\right(s-a\left)\right(s-b\left)\right(s-c\left)\right]}$

$=\frac{\sqrt{s}\sqrt{s}(3s-2s)}{\sqrt{s}(s-a)(s-b)(s-c)]}=\frac{s^2}{S}$

and Cot A + cot B + cos C

$=\frac{cosA}{sinA}+\frac{CosB}{sinB}+\frac{cosC}{sinC}$

$=\frac{(b^2+c^2-a^2)/2bc}{2S/bc}+\frac{(c^2+a^2-b^2)/2ca}{2S/ca}+\frac{(a^2+b^2-c^2)/2ab}{2S/ab}$

$=\left(1/4s\right)[b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2]$

$=\frac{a^2+b^2+c^2}{4S}.$ ...........(I)

Hence R.H.S.

$=\frac{s^2}{S}\times \frac{4S}{a^2+b^2+c^2}=\frac{4s^2}{a^2+b^2+c^2}=\frac{(a+b+c{)}^2}{a^2+b^2+c^2}$.