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?

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    $\begingroup$ Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ Jul 9, 2013 at 3:13
  • $\begingroup$ Recently, this edit happened where a user attempted to convert the units being used from degrees to radians. Needless to say this is a non-improvement and is potentially confounding for the OP. I wish I could send a gentle reminder to the two users who approved the request to use their approvals with more discretion. :S $\endgroup$
    – rschwieb
    Aug 8, 2013 at 15:10

4 Answers 4


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 $$\tan x=\dfrac{\sin x}{\cos x} = \dfrac{\sqrt 7/4}{-3/4} = -\left(\dfrac{\sqrt 7}{3}\right)$$ $$\csc x=\dfrac1{\sin x}= \dfrac{1}{\sqrt 7/4} = \dfrac 4{\sqrt 7} $$

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    $\begingroup$ Deserves another upvote! +1 $\endgroup$
    – Amzoti
    Jul 9, 2013 at 3:45

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 $\csc x=\frac{4}{\sqrt{7}}$.


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 $\tan x=\frac{\sin x}{\cos x}=...$


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.


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