Question

If ${\tan }^2\theta -4\sqrt{3}\tan \theta +3=0$, then find $\tan \theta$

Answer 1

${\tan }^2\theta -4\sqrt{3}\tan \theta +3=0$

$\tan \theta =\frac{4\sqrt{3}\pm \sqrt{(4{\sqrt{3}}^2-4(1\left)\right(3)}}{2}$

$=\frac{4\sqrt{3}\pm \sqrt{16\left(3\right)-4\left(3\right)}}{2}$

$=\frac{4\sqrt{3}\pm \sqrt{36}}{2}$

$=\frac{4\sqrt{3}\pm 6}{2}$

$=2\sqrt{3}\pm 3$

$\Rightarrow \tan \theta =2\sqrt{3}\pm 3$.