$A+B+C=2\pi$, prove determinant equals to zero.

Question

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?

Answer 1

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.