# How do you find the value of cot 300^@?

Dec 9, 2014

To find the value of $\cot 300$, you will first need to write the angle as the sum of difference of two angles, one of which is either ${90}^{\circ} , {180}^{\circ} , {270}^{\circ} \mathmr{and} {360}^{\circ}$.

Note: Remember that when you write it with ${90}^{\circ} \mathmr{and} {270}^{\circ}$, the fuction will change to it's co-function, in this case, to $\tan$.

Let's first look at the two easiest ways to write this:

$\cot {300}^{\circ} = \cot {\left(270 + 30\right)}^{\circ}$

and

$\cot {300}^{\circ} = \cot {\left(360 - 60\right)}^{\circ}$

An important thing to remember is in which quadrants will a trigonometric function be positive. Here's an illustration: Here,
A stands for all.
S stands for sin.
T stands for tan.
C stands for cos.

This means that
all fuctions are positive in the first quadrant,
the sin function and it's co-function csc are positive in the second quadrant,
the tan function and it's co-function cot are positive in the third quadrant,
the cos function and it's co-function sec are positive in the fourth quadrant.

One way to remember this arrangement is to recite the sentence:

$A$ll $S$tudents $T$ake $C$alculus.
This tells us which function would be positive in which quadrant.

I personally like to use the sentence
$A$ll $S$cience $T$eachers are $C$razy.

So, let's solve using the first equation.

$\cot {300}^{\circ} = \cot {\left(270 + 30\right)}^{\circ}$

The angle is greater than ${270}^{\circ}$ and thus lies in the fourth quadrant. $\tan$ and $\cot$ are not positive here, i.e., they are negative.

Also, since you've used ${270}^{\circ}$, you need to change it to $\tan$.

$\cot {300}^{\circ} = \cot {\left(270 + 30\right)}^{\circ} = - \tan {30}^{\circ}$

$\cot {300}^{\circ} = - \frac{1}{\sqrt{3}}$

Now, let's solve the second equation.

$\cot {300}^{\circ} = \cot {\left(360 - 60\right)}^{\circ}$

Here, the angle is expressed with ${360}^{\circ}$, you must keep the function as $\cot$ itself. Again, the angle lies in the fourth quadrant, which means it is negative.

$\cot {300}^{\circ} = \cot {\left(360 - 60\right)}^{\circ} = - \cot {60}^{\circ}$

$\cot {300}^{\circ} = - \frac{1}{\sqrt{3}}$