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?
 A: 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 $$\tan x=\dfrac{\sin x}{\cos x} = \dfrac{\sqrt 7/4}{-3/4} = -\left(\dfrac{\sqrt 7}{3}\right)$$  $$\csc x=\dfrac1{\sin x}= \dfrac{1}{\sqrt 7/4} = \dfrac 4{\sqrt 7} $$
A: 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 $\csc x=\frac{4}{\sqrt{7}}$.
A: As $90^\circ< x<180^\circ,\tan x <0,\sin x>0$
So, $\sin x=+\sqrt{1-\cos^2x}=...$
$\csc x=\frac1{\sin x}= ... $
and $\tan x=\frac{\sin x}{\cos x}=...$
A: You should draw a triangle!
They tell you that you're working in the second quadrant, so draw a reference triangle and label the base angle $x$.    
Since cosine is the ratio of adjacent ("over"/) to hypotenuse, we know that the length of the adjacent side is the numerator $ = 3$.
The length of the hypotenuse is the denominator $= 4$... But where does the "-" go?? 
The hypotenuse is always positive. That, and the fact that you are in the left quadrant (<=> the x-coordinate is negative) should tell you that the adjacent side should really be labeled $-3$. This will be important later, as the tangent requires the adjacent side. 
Now use the Pythagorean Theorem to find the length of the opposite side. It's $\sqrt 7$.
Then you can find tangent (opposite over adjacent), sine (opposite over hypotenuse), etc. using SOHCAHTOA.   
