Answer
$135^{\mathrm{o}},\quad 315^{\mathrm{o}}$
Work Step by Step
Since $\cot\theta$ is negative, $\theta$ must lie in quadrants II or IV.
Since the absolute value of $\cot\theta$ is 1,
browsing through the table: Function Values of Special Angles,
the reference angle, $\theta^{\prime}$ must be $45^{\mathrm{o}}$.
Now, from
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
In quadrant II
$\theta'=180^{o}-\theta$
so
$\theta=180^{o}-\theta^{\prime}=180^{o}-45^{o}=135^{\mathrm{o}}$
In quadrant IV,
$\theta'=360^{o}-\theta$
so
$\theta=360^{o}-\theta^{\prime}=360^{o}-45^{o}=315^{\mathrm{o}}$.
