Question

Solve: $\tan x=\frac{\tan {60}^o-\tan {30}^o}{1+\tan {60}^o\tan {30}^o}$

Answer 1

$tanx=\frac{tan{60}^{\circ }-tan{30}^{\circ }}{1+tan{60}^{\circ }tan{30}^{\circ }}$

$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}$

$\Rightarrow \frac{3-1}{\sqrt{3}\left(2\right)}=\frac{1}{\sqrt{3}}$

$tanx=\frac{1}{\sqrt{3}}$

$x={30}^{\circ }$