If csc theta =-2 and theta lies in quadrant 3, find tan theta

Answer 1

Since $\csc\theta$ is the reciprocal of $\sin\theta$, and $\csc\theta = -2$, we have $\sin\theta = -\frac{1}{2}$.

The sine function is negative in Quadrant III, so we know that $\sin\theta < 0$. Since $\sin\theta < 0$ and $\sin\theta = -\frac{1}{2}$, we can draw a right triangle in Quadrant III with opposite side $1$ and hypotenuse $2$. Since the opposite side is $1$, the adjacent side must be $\sqrt{2^2 - 1^2} = \sqrt{3}$.

Therefore, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-1/2}{-\sqrt{3}/2} = \boxed{\frac{1}{\sqrt{3}}}$.