Q.In $\triangle ABC$, prove that $\dfrac{\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}}{\cot A + \cot B + \cot C} = \dfrac{(a+b+c)^2}{a^2+b^2+c^2}$.

Half-Angle Formulas

The cosine rule gives $\cos A$ in terms of the three sides, but many later results — Heron's formula, and every inradius/exradius theorem — turn out to be cleanest when expressed through the half-angles $\frac A2,\frac B2,\frac C2$ instead. Writing $s=\frac{a+b+c}2$ (the semi-perimeter), the half-angle formulas state:

and cyclically for $B,C$.

Why express things in $s-a,s-b,s-c$?

The quantities $s-a,s-b,s-c$ are always positive for a genuine triangle (triangle inequality), and they are exactly the tangent lengths from each vertex to the incircle/excircles — so once you have the half-angle formulas in this form, the incircle and excircle theorems become almost immediate substitutions rather than fresh derivations.

Where they come from

Start from the cosine rule $\cos A=\frac{b^2+c^2-a^2}{2bc}$ and the double-angle identities $\cos A=1-2\sin^2\frac A2$ and $\cos A=2\cos^2\frac A2-1$. Substituting and factoring $a^2-(b-c)^2=(a-b+c)(a+b-c)$ and $(b+c)^2-a^2=(b+c-a)(b+c+a)$ — both differences of squares — converts everything into products of $(s-a),(s-b),(s-c),s$, after dividing by $2$ throughout.

A common trap

It's tempting to take $\sin\frac A2=\pm\sqrt{\cdots}$, but because $0<A<180^\circ$ forces $0<\frac A2<90^\circ$, both $\sin\frac A2$ and $\cos\frac A2$ must be positive — the negative root is never physically valid here, unlike in some other trigonometric contexts.

Worked micro-example …