Question

If $f\left(x\right)=\frac{1+\tan x}{1-\tan x}$ then $f\left(\frac{\pi }{6}\right)$

Answer 1

$f\left(\frac{\pi }{6}\right)=\frac{1+\tan \frac{\pi }{6}}{1-\tan \frac{\pi }{6}}$

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

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

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

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

$=\frac{4+2\sqrt{3}}{2}$

$=2+\sqrt{3}$