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.
 A: I have got this here for your determinant: $$\tan \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right)-\cot \left(\frac{A}{2}\right)
   \tan \left(\frac{B}{2}\right)-\tan \left(\frac{A}{2}\right) \cot
   \left(\frac{C}{2}\right)+\cot \left(\frac{A}{2}\right) \tan \left(\frac{C}{2}\right)+\tan
   \left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)-\cot \left(\frac{B}{2}\right) \tan
   \left(\frac{C}{2}\right)=0$$
With $$C=\pi-A-B$$ we get
$$-\frac{8 (\sin (A-B)+\sin (2 A+B)-\sin (A+2 B))}{-\sin (2 A+2 B)+\sin (2 A)+\sin (2 B)}=0$$
You have to solve 
$$\sin(A-B)+\sin(2A+B)=\sin(A+2B)$$
One solution is given by $$A=B$$
A: Hint:
Applying $C_2'=C_2-C_1,C_3=C_3-C_1$
$$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix} =\begin{vmatrix}1& 1-1& 1-1\\
\displaystyle \dfrac1p & \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\
q+r & r+p-(q+r)& p+q-(q+r)\end{vmatrix}$$
$$=1\cdot\begin{vmatrix} 
\displaystyle  \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\r+p-(q+r)& p+q-(q+r)\end{vmatrix} $$
$$=\dfrac{(p-q)(p-r)}{pq}-\dfrac{(p-r)(p-q)}{pr}=?$$
A: With transformation $R_3 \longrightarrow R_3- \left(\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{B}{2}\right)+\tan\left(\dfrac{C}{2}\right)\right)R_1$
we get $\begin{vmatrix}
 1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} &  \displaystyle \cot \frac{C}{2}  \\\\ 
 \displaystyle -\tan\frac{A}{2} &\;\;\displaystyle -\tan \frac{B}{2} & \;\;\displaystyle -\tan \frac{C}{2}\end{vmatrix}$
Now factor out $(-1)$ from Row $3$ and multiply the columns by $\displaystyle \tan\frac{A}{2}, \displaystyle \tan\frac{B}{2}, \displaystyle \tan\frac{C}{2}$ respectively to obtain
$-\begin{vmatrix}
 \displaystyle \tan\frac{A}{2} &\;\;\displaystyle \tan \frac{B}{2} & \;\;\displaystyle\tan \frac{C}{2}\;\;\\\\
1 & \;\;1\;\;&\;\; 1 \\\\ 
 \displaystyle \tan^2\frac{A}{2} &\;\;\displaystyle \tan^2 \frac{B}{2} & \;\;\displaystyle\tan^2 \frac{C}{2}\end{vmatrix}$ which can be simplified to the Vandermonde Determinant and hence it evaluates to $\displaystyle \left(\tan\frac{A}{2} - \tan\frac{B}{2}\right)\left(\tan\frac{B}{2} - \tan\frac{C}{2}\right)\left(\tan\frac{C}{2} - \tan\frac{A}{2}\right)$
