## Trigonometry (11th Edition) Clone

$135^{\mathrm{o}},\quad 315^{\mathrm{o}}$
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 $\theta=360^{o}-\theta^{\prime}=360^{o}-45^{o}=315^{\mathrm{o}}$.