Exact arctangent of product of tangents > Calculate $x$, if $$\tan(x)=\tan9\tan69\tan33$$ (Using sexagesimal degrees) Since $\tan3x=\tan(60-x)\tan x \tan(60+x)$: \begin{align*} \tan27&=\tan69\tan9\tan51\\\ \implies\tan27\tan39&=\tan69\tan9 \end{align*} So the problem is equivalent to calculating $x$ in $$\tan(x)=\tan27\tan33\tan39$$ But thats all my progress so far. Interestingly enough, the answer is $x=15$. Is there some way tu constructively solve the equation? If not, a straightforward proof of $\tan(15)=\tan9\tan69\tan33$ would be nice too.

$$ \tan(x)=\tan(9)\tan(69)\tan(33) $$ $$ \tan(x) = \frac{\tan(39)\tan(3)}{\tan(9)} $$ Thus, $$ \tan^2(x) = \tan(3)\tan(33)\tan(39)\tan(69) $$ $$ \tan^2(x) = \tan(3)\tan(3-36)\tan(3-72)\tan(3+36) $$ $$ \tan^2(x)\tan(75) = \tan(3-72)\tan(3-36)\tan(3)\tan(3+36)\tan(3+72) $$

I will now show that ,

$$ \tan(5x)= \tan(x-72)\tan(x-36)\tan(x)\tan(x+36)\tan(x+72) $$

Let $z=\cos(x)+i\sin(x)$, and $ \omega = \cos(36)+i\sin(36) $ Then, $$ \tan(x) = -\frac{i(z^2-1)}{2(z^2+1)} $$ Similarly we can get $\tan(x-36),\tan(x-72),\tan(x+36),\tan(x+72) $ and then multiply . We very easily get the above identity. So, we have $$ \tan^2(x)=\tan^2(15) \implies x=15 $$