Question

Find the value of $\tan \frac{13\pi }{12}$

Answer 1

$\tan \frac{13\pi }{12}$ = $\tan (\pi +\frac{\pi }{12})=\tan \left(\frac{\pi }{12}\right)$

=$\tan (\frac{\pi }{4}-\frac{\pi }{6})=\frac{\tan \frac{\pi }{4}-\tan \frac{\pi }{6}}{1+\tan \frac{\pi }{4}\tan \frac{\pi }{6}}$

$\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

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

$\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}{)}^2-(1{)}^2}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=\frac{2(2-\sqrt{3})}{2}=2-\sqrt{3}$

$\therefore$ The value of $\tan \left(\frac{13\pi }{12}\right)\text{is}2-\sqrt{3}.$