Find all trig ratios of $\cot \beta = -\frac13$ Find all the trig ratios where $\cot \beta = -\dfrac13$, and $\pi < \beta < 2\pi$.
I understand how to do this type of problem for sine and cosine, but with tangent and cotangent I don't understand how to do it.
The answers for some of the trig ratios are $\sin \beta = -\dfrac1{\sqrt{10}}$ and $\cos \beta = \dfrac{-3}{\sqrt{10}}$.
In particular, how do we arrive at the $\sqrt{10}$?
 A: Since, It is provided that $\beta$ is either in quadrant III or IV $(\pi < \beta < 2\pi)$, but its value is negative so it must be in quadrant IV $\therefore$ we will add sign accordingly.
Given that, $\cot \beta = -\dfrac13$,
We know that, $\cot^2\beta+1=\csc^2\beta$  
Which gives, $\csc\beta=-\sqrt{1+\dfrac{1}{9}}=-\sqrt{\dfrac{10}{9}}=-\dfrac{\sqrt{10}}{3}$
Since, $\sin\beta=\dfrac{1}{\csc\beta}\implies\sin\beta=-\dfrac{3}{\sqrt{10}}$ And,
$\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-\dfrac{9}{10}}=\dfrac1{\sqrt{10}}$
A: $\tan \beta = -3$.  Then you can use $1+\tan^2 \beta=\sec^2 \beta$.  
For cotangent to be negative you must be in quadrant II or IV; but from the given information you are in quadrant IV.  Use this to decide which square root to take to get $\sec \beta$.
From the value of $\sec \beta$, you can get $\cos \beta$.
A: We have $\dfrac{\cos\beta}{\sin\beta}=-\dfrac13$
$\iff \dfrac{\cos\beta}1=\dfrac{\sin\beta}{-3}=\pm\sqrt{\dfrac{\cos^2\beta+\sin^2\beta}{1^2+(-3)^2}}=\pm\dfrac1{\sqrt{10}}$
Now as $\pi<\beta<2\pi$ and $\cot\beta=-\dfrac13<0,\dfrac{3\pi}2\le\beta<2\pi$
$\implies(i)\sin\beta<0$ and   $\implies\sin\beta=-\dfrac3{\sqrt{10}}$
$(ii)\cos\beta>0\implies\cos\beta=\dfrac1{\sqrt{10}}$
$\tan\beta=1/\cot\beta$ and so on
A: Hint
Try drawing a unit circle that has been scaled by a factor or $r$. Then draw the angle ($\beta$) in the correct quadrant, and draw the reference triangle. Label the known lengths and solve for $r$. You can now "read" all of the trigonometric ratios from the reference triangle.

A: Imagine drawing the triangle on the on a coordinate circle.  You know that either the adjacent (x) side or the opposite (y) side must be negative (but not both).  You also know it is in quadrants III or IV.  So we must be in quadrant 4 because in that quadrant, $y$ is negative and $x$ is not.  So construct a triangle in that quadrant with an adjacent side of 1 and opposite side of $-3$.  Using the pythagorean theorem, you can see that the hypotenuse will be $\sqrt{1^2+(-3)^2}=\sqrt{10}$
EDIT: If $\cot{\beta}=-\frac{1}{3}$, I don't think $\sin{\beta}=-\frac{1}{\sqrt{10}}$.  
I think $\sin{\beta}=-\frac{3}{\sqrt{10}}$ and $\cos{\beta}=\frac{1}{\sqrt{10}}$
Let me check my work...
