For a circle with angles $\alpha$, $\beta$, $\gamma$, inradius $r$, and semiperimeter $p$, prove $\tan(\alpha/2)\tan(\beta/2)\tan(\gamma/2)=r/p$

Question

Prove that for every triangle with angles $\alpha$, $\beta$ and $\gamma$ and the radius of the inscribed circle $r$ and the semipermeter $p$ is true that $\tan\frac{\alpha}{2}*\tan\frac{\beta}{2}*\tan\frac{\gamma}{2}=\frac{r}{p}$.

I have no idea even on how to start and my experience with trigonometry isn't much so I really need help.

Answer 1

Hint: try the half angle identity for the tangent

$\tan(\frac{a}{2})=\pm\sqrt{\frac{1-\cos(a)}{1+\cos(a)}}=\frac{1-\cos(a)}{\sin(a)}=\frac{\sin(a)}{1+\cos(a)}$

Answer 2

Use https://en.m.wikibooks.org/wiki/Trigonometry/Solving_triangles_by_half-angle_formulae

$$\tan\dfrac A2=\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}}$$

Finally $$\triangle =\sqrt{p(p-a)(p-b)(p-c)}=rp$$

See https://proofwiki.org/wiki/Area_of_Triangle_in_Terms_of_Inradius