Product of tangents -part 1

Using the identity $\tan (3x) = \tan(60^{\circ} - x) \tan (x) \tan (60^{\circ} + x)$ , we have

$\tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ} = \frac{\left(\tan 48^{\circ} \tan 12^{\circ} \tan 72^{\circ} \right)\tan 24^{\circ}}{\tan 72^{\circ} } = \frac{\left(\tan 36^{\circ} \right)\tan 24^{\circ}}{\tan 72^{\circ} } = \frac{\tan 36^{\circ}\tan 24^{\circ}\tan 84^{\circ}}{\tan 72^{\circ}\tan 84^{\circ} } = \frac{\tan 72^{\circ}}{\tan 72^{\circ}\tan 84^{\circ} }=\frac{1}{\tan 84^{\circ} }=\tan 6^{\circ}$

As $0<x<90$ and $\tan x =\tan 6^{\circ}$ , $\boxed{x=6}$ .

Using the identity ( equation (35) ),

$\prod_{k=1}^{\lfloor (n-1)/2 \rfloor} \tan \left(\frac {k\pi}n \right) = \begin{cases} \sqrt n & \text{if }n \text{ is odd} \\ 1 & \text{if }n \text{ is even} \end{cases}$

we have:

$\begin{aligned} \blue{\tan \frac \pi{15}} \blue{\tan \frac {2\pi}{15}} \tan \frac \pi 5 \blue{\tan \frac {4\pi}{15}} \tan \frac \pi 3 \tan \frac {2\pi}5 \tan \frac {7\pi}{15} & = \sqrt {15} \\ \blue{\tan 12^\circ \tan 24^\circ \tan 48^\circ} \tan \frac \pi 3 \red{\tan \frac \pi 5 \tan \frac {2\pi}5} \cot \left(\frac \pi 2 - \frac {7\pi}{15}\right) & = \sqrt {15} & \small \red{\text{Note that }\prod_{k=1}^2 \tan \frac {k\pi}5 = \sqrt 5} \\ \blue{\tan 12^\circ \tan 24^\circ \tan 48^\circ} \cdot \sqrt 3 \cdot \red{\sqrt 5} \cdot \cot \frac \pi{30} & = \sqrt {15} \\ \implies \tan 12^\circ \tan 24^\circ \tan 48^\circ & = \tan \frac \pi{30} = \tan 6^\circ \end{aligned}$

Therefore $x = \boxed 6$ .