Question

Find \tan\left(\frac{17\pi}{12}\right)tan( 12 17π ​ )tangent, left parenthesis, start fraction, 17, pi, divided by, 12, end fraction, right parenthesis exactly using an angle addition or subtraction formula.

Answer 1

One way to do this is to notice

$\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4$

Then

$\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}$

We have

$\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

and since $\tan x$ has a period of $\pi$,

$\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1$

and so

$\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3$