Question

Prove
$tan{75}^o=2+\sqrt{3}$

Answer 1

$\tan 75°=2+\sqrt{3}$

$\tan (45°+30°)$

$=\frac{\tan 45°+\tan 30°}{1-\tan 45°\tan 30°}$

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

$=\frac{{(1+\frac{1}{\sqrt{3}})}^2}{1-\frac{1}{\sqrt{3}}}$

$=\frac{1+\frac{1}{3}+\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$=2+\sqrt{3}$

Hence proof