if $a,b,c$ and area of triangle $ABC$ are rational then prove that $\tan \frac{B}{2},\tan \frac{C}{2}$ are rational

Question

consider the following statement about the triangle $ABC$

$(a)$ the sides length $a,b,c$ and the area of triangle $ABC$ are rational

$(b)$ side length $a\;$ and $ \displaystyle \tan \left(\frac{B}{2}\right),\tan \left(\frac{C}{2}\right)$ are rational

$(c)$ the side $a$ and $\sin A, \sin B,\sin C$ are rational

then prove that $(a)\Rightarrow (b)\Rightarrow (c)\Rightarrow (a).$

Attempt if $a,b,c$ are rational and area of triangle $\displaystyle ABC = 0.5 ab \sin C$ are rational

and $\displaystyle \tan B = \frac{1-\tan^2 \frac{B}{2}}{1+\tan^2 \frac{B}{2}}$ and $\displaystyle \tan C = \frac{1-\tan^2 \frac{C}{2}}{1+\tan^2 \frac{C}{2}}$

how can prove that $\displaystyle \tan \frac{B}{2}\;, \tan \frac{C}{2}$ are rational

could some help me with this, thanks

Answer 1

Hints:

---

For $(a) \Rightarrow (b) $, we know that the semiperimeter $s $ is rational. Thus use the identities $$\tan \frac {B}{2} =\frac {\Delta}{s (s-b)} \text { and } \tan \frac {C}{2} =\frac {\Delta}{s (s-c)} $$ where $\Delta $ is the area of $\triangle ABC$ to conclude.

---

For $(b) \Rightarrow (c) $, we know that $$\sin B =\frac {2\tan B/2}{1+\tan^2 B/2}$$ $$\sin C=\frac {2\tan C/2}{1+\tan^2 C/2} $$ and $$\tan A/2 = \tan (\frac {\pi}{2} -\frac {B+C}{2}) $$ to conclude.

---

For $(c) \Rightarrow (a) $, use the law of sines to conclude.

Hope it helps.