Answer
$45^{o},\quad 225^{o}$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quad.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
Tangent is positive in quadrants I and III .
Browsing through: Function Values of Special Angles,
we find that $\tan 45^{o}=1$, so the reference angle is $45^{o}$
In quadrant I, $\theta=45^{o}$
In quadrant III, $\theta^{\prime}=\theta-180^{o}$ so
$\theta=180^{o}+\theta^{\prime}=180^{o}+45^{o}=225^{o}$
