Proof:\begin{equation} \text{det} \begin{pmatrix} 1 & 1 & 1 \\ \tan A & \tan B & \tan C \\ \tan 2A & \tan 2B & \tan 2C \end{pmatrix}=0 \end{equation} where A+B+C=$2\pi$.
This problem of course can be prove by expand this determinant explicitly and calculate it.I wonder if there exists a more convenient way.
This determinant has a representation as $ \sum_{cyc}\{\tan B\tan 2C-\tan C\tan 2B\} $
Are you sure that it's true?
Take $A=100^{\circ},$ $B=120^{\circ}$ and $C=140^{\circ}.$
By the way, the following reasoning can be interesting.
Let $\tan A=a$, $\tan B=b$ and $\tan C=c$.
Thus, $a+b+c=abc$ and $$\Delta=\sum_{cyc}\left(\frac{2bc}{1-c^2}-\frac{2bc}{1-b^2}\right)=\frac{2}{\prod\limits_{cyc}(1-a^2)}\sum_{cyc}bc(c^2-b^2)(1-a^2)=$$ $$=\frac{2}{\prod\limits_{cyc}(1-a^2)}\sum_{cyc}(a^3c-a^3b+a^3c^2b-a^3b^2c)=$$ $$= \frac{2(a-b)(b-c)(c-a)(a+b+c+abc)}{\prod\limits_{cyc}(1-a^2)}.$$