$A+B+C=2\pi$, prove determinant equals to zero. Given $A$, $B$, $C$ which satisfy $A+B+C=2\pi$, is there an ingenious method to prove that
$$
\det\begin{pmatrix}
    1 & 1 & 1 \\
    \tan A & \tan B & \tan C \\
    \tan 2A & \tan 2B & \tan 2C
\end{pmatrix}=0
$$? By column transformation we have
$$
\det\begin{pmatrix}
    1 & 1 & 1 \\
    \tan A & \tan B & \tan C \\
    \tan 2A & \tan 2B & \tan 2C
\end{pmatrix}=\det\begin{pmatrix}
\tan B-\tan A & \tan C-\tan B \\
\tan 2B-\tan 2A & \tan 2C-\tan 2B
\end{pmatrix}.
$$
Substitute $C=2\pi—A-B$ and the rest computation seems very complicated! Any hint?

Well here is my working of the problem. I must have made a sign mistake somewhere, however if I have not, then the equation only holds under special cases. Feedback welcome.
Let $a=\tan A, b=\tan B,c=\tan C$ then
$$\tan 2A=\frac{2a}{1-a^2}$$
etc.
So the matrix is
$$X=\begin{vmatrix}
1&1&1\\
a&b&c\\
\frac{2a}{1-a^2}&\frac{2b}{1-b^2}&\frac{2c}{1-c^2}\\
\end{vmatrix}$$
Multiplying the columns the question is transformed into
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X=\begin{vmatrix}
1-a^2&1-b^2&1-c^2\\
a-a^3&b-b^3&c-c^3\\
a&b&c\\
\end{vmatrix}$$
Which is the same as
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X=\begin{vmatrix}
a&b&c\\
a^2-1&b^2-1&c^2-1\\
a^3&b^3&c^3\\
\end{vmatrix}$$
Now we calculate,
$$\begin{vmatrix}
a&b&c\\
a^2-1&b^2-1&c^2-1\\
a^3&b^3&c^3\\
\end{vmatrix}=\begin{vmatrix}
a&b&c\\
a^2&b^2&c^2\\
a^3&b^3&c^3\\
\end{vmatrix}-\begin{vmatrix}
a&b&c\\
1&1&1\\
a^3&b^3&c^3\\
\end{vmatrix}$$
$$=abc\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\
\end{vmatrix}+\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3\\
\end{vmatrix}$$
Now by assumption, $a+b+c=abc$
so we have
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X$$
$$=(a+b+c)\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\
\end{vmatrix}+\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3\\
\end{vmatrix}$$
Note however that the difference of these two matrices is zero,
$$(a+b+c)\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\
\end{vmatrix}-\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3\\
\end{vmatrix}$$
$$=\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2(b+c)&b^2(a+c)&c^2(a+b)\\
\end{vmatrix}=0$$
because
$$bc^2(a+b)-cb^2(a+c)=abc(c-b)$$
$$ca^2(b+c)-ac^2(a+b)=abc(a-c)$$
$$ab^2(a+c)-ba^2(b+c)=abc(b-a)$$
Therefore,
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X=
2(a+b+c)\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\\end{vmatrix}$$
So
$$X=\frac{4abc(a-b)(b-c)(c-a)}{(1-a^2)(1-b^2)(1-c^2)}$$
$$=\frac{1}{2}\tan 2A \tan 2B\tan 2C(\tan A-\tan B)(\tan B-\tan C)(\tan C-\tan A)$$
And this $X=0$ happens only in special cases.
