Precalculus (6th Edition)

$\{45^{o},135^{o},225^{o},315^{o}\}$
Substituting t$=\tan\theta$ ,the equation becomes $t-\displaystyle \frac{1}{t}=0,\qquad t\neq 0$ $t-\displaystyle \frac{1}{t}=0\quad/\times t$ $t^{2}-1=0$ ... difference of squares ... $(t+1)(t-1)=0$ ... apply the zero factor principle... $t=1\quad$or$\quad t=-1$ $\tan\theta=1\quad$or$\quad\tan\theta=-1$ $\theta=45^{o},225^{o}\quad$or$\quad \theta=135^{o},315^{o}$ Solution set= $\{45^{o},135^{o},225^{o},315^{o}\}$