TITLE: Given that $\cos x =-3/4$ and $90^\circ<x<180^\circ$, find $\tan x$ and $\csc x$
QUESTION [5 upvotes]: Given that $\;\cos x =-\frac{3}{4}\,$ and $\,90^\circ<x<180^\circ,\,$ find $\,\tan x\,$ and $\,\csc x.$
This question is quite unusual from the rest of the questions in the chapter, can someone please explain how this question is solved? I tried Pythagorean Theorem, but no luck. Is it possible to teach me how to use the circle diagram?

REPLY [2 votes]: We will figure out signs later. Draw a right-angled triangle. Call one of the small angles $t$. We want the cosine of $t$ to be $\frac{3}{4}$ (not a typo). So we want the hypotenuse to be $4$ (please write it in) and the adjacent side to be $3$. By the Pythagorean Theorem, the opposite side is $\sqrt{16-9}$, that is, $\sqrt{7}$.
Then $\tan t=\frac{\sqrt{7}}{3}$. And $\csc t=\frac{1}{\sin t}=\frac{4}{\sqrt{7}}$.
Now back to $x$. The trigonometric functions of $x$ will have the same absolute value as the corresponding trigonometric functions of $t$, but maybe different signs.
In the second quadrant, $\tan$ is negative. So $\tan x=-\frac{\sqrt{7}}{3}$.
In the second quadrant, $\sin$ is positive. So $\csc x=\frac{4}{\sqrt{7}}$.