## Trigonometry (11th Edition) Clone

There are 4 values of $\theta$ in need to find, which is {$60^\circ, 135^\circ, 240^\circ, 315^\circ$}.
$$\tan\theta+1=\sqrt3+\sqrt3\cot\theta$$ $$\tan\theta+1=\sqrt3+\sqrt3\frac{1}{\tan\theta}$$ $$\tan\theta-\frac{\sqrt 3}{\tan\theta}+1-\sqrt 3=0$$ $$\frac{\tan^2\theta-\sqrt 3+(1-\sqrt 3)\tan\theta}{\tan\theta}=0$$ Divide both sides by $\tan\theta$, we have $$\tan^2\theta-\sqrt 3+(1-\sqrt 3)\tan\theta=0$$ Here you can solve with calculator or do something like this: $$\tan^2\theta-\sqrt3+\tan\theta-\sqrt 3\tan\theta=0$$ $$(\tan^2\theta+\tan\theta)+(-\sqrt 3-\sqrt 3\tan\theta)=0$$ $$\tan\theta(\tan\theta+1)-\sqrt 3(\tan\theta+1)=0$$ $$(\tan\theta+1)(\tan\theta-\sqrt 3)=0$$ $$\tan\theta=-1$$ or $$\tan\theta=\sqrt 3$$ *For $\tan\theta=-1$ $\tan\theta$ is negative, which means the angles must lie in quadrant II and IV. Also, the reference angle would be $45^\circ$. 2 angles satisfy the above requirements which is {$135^\circ, 315^\circ$}. *For $\tan\theta=\sqrt 3$ $\tan\theta$ is positive, which means the angles must lie in quadrant I and III. Also, the reference angle would be $60^\circ$. 1 angle satisfy the above requirements which is {$60^\circ, 240^\circ$}. Overall, there are 4 values of $\theta$ in need to find, which is {$60^\circ, 135^\circ, 240^\circ, 315^\circ$}.