<s>[INST] Given that $\cos x =-3/4$ and $90^\circGiven 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?
 [/INST] The cosine of an angle corresponds to the $x$-coordinate in the unit circle, and the sine of an angle cooresponds to its $y$-coordinate on the unit circle.
Note that $\,\cos = -\frac 34 < 0\,$ if and only if the angle $x$ terminates in either the second or third quadrant, where the angle $x$ is measured with respect to the positive $x$-axis. 
Since we are given that $\,90 \lt x \lt 180,\,$ we know that the angle $x$ terminates in the second quadrant. So $\sin x > 0$.
So, $\,\tan x = \dfrac{\sin x}{\cos x} <0,\;$ and $\,\csc x = \dfrac{1}{\sin x}>0$.
Now, we know that by the Pythagorean Theorem as it relates to trigonometry identities, $${\bf \sin^2 x + \cos^2 x = 1} $$ $$ \begin{align} \iff \sin^2 x & = 1 -\cos^2 x \\ \\ \implies  \sin x & = \pm \sqrt{1 -\cos^2 x} \\ \\ & = \pm \sqrt{1 - \left(\dfrac {-3}{4}\right)^2} \\ \\ & = \pm \sqrt{\frac 7{16}} \\ \\ & = \pm \frac{\sqrt 7}{4}\end{align}$$ Since we know that in the second quadrant, $\sin x > 0$, we take the positive root: $$\sin x = \frac{\sqrt 7}{4}$$
and so you have all you need to compute $$\tan x=\dfrac{\sin x}{\cos x} = \dfrac{\sqrt 7/4}{-3/4} = -\left(\dfrac{\sqrt 7}{3}\right)$$  $$\csc x=\dfrac1{\sin x}= \dfrac{1}{\sqrt 7/4} = \dfrac 4{\sqrt 7} $$
</s>