<s>[INST] prove that the triangle is isosceles 
In a $\triangle ABC$, If 
$\begin{vmatrix}
 1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} &  \displaystyle \cot \frac{C}{2}  \\\\ 
 \displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2}  & \;\;\displaystyle\tan \frac{A}{2}+\tan \frac{B}{2}\end{vmatrix}=0$
Then prove that the triangle is isosceles.

Try: Let $\displaystyle \tan \frac{A}{2}=p\;\;,\tan \frac{B}{2}=q\;\;,\tan \frac{C}{2}=r$
Using $$\tan\bigg(\frac{A}{2}+\frac{B}{2}\bigg)=\tan\bigg(\frac{\pi}{2}-C\bigg)$$
So $$\sum \tan\frac{A}{2}\tan\frac{B}{2}=1\Rightarrow pq+qr+rp=1$$
So $$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
p(q+r) & q(r+p)& r(p+q)\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
1-qr & 1-rp& 1-pq\end{vmatrix}=0$$
So  $$-\begin{vmatrix}p^2& q^2& r^2\\
\displaystyle p & \displaystyle q & \displaystyle r \\
1 & 1& 1\end{vmatrix}=0$$
So we have $(p-q)(q-r)(r-p)=0.$
So either $p=q$ or $q=r$ and $r=p.$
Could some help me some short way to solve it?  Thanks.
 [/INST] Hint:
Applying $C_2'=C_2-C_1,C_3=C_3-C_1$
$$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix} =\begin{vmatrix}1& 1-1& 1-1\\
\displaystyle \dfrac1p & \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\
q+r & r+p-(q+r)& p+q-(q+r)\end{vmatrix}$$
$$=1\cdot\begin{vmatrix} 
\displaystyle  \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\r+p-(q+r)& p+q-(q+r)\end{vmatrix} $$
$$=\dfrac{(p-q)(p-r)}{pq}-\dfrac{(p-r)(p-q)}{pr}=?$$
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