Question

$tan\frac{13\pi }{12}=2-\sqrt{n}$. Then $n=$

Answer 1

$tan\frac{13\pi }{12}=tan(\pi +\frac{\pi }{12})=tan\frac{\pi }{12}=tan(\frac{\pi }{4}-\frac{\pi }{6})$
$=\frac{tan\frac{\pi }{4}-tan\frac{\pi }{6}}{1+tan\frac{\pi }{4}tan\frac{\pi }{6}}$
$=\frac{1-\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=2-\sqrt{3}$