# Are cotangent and negative tangent graphs the same?

Sep 21, 2015

No. See explanation.

#### Explanation:

Let's look at the definitions in terms of sines and cosines (because tangents and cotangents are just ratios of sines and cosines).

Cotangent of x is:
$\cot x = \cos \frac{x}{\sin} x$

and negative tangent of x is:
$- \tan x = - \sin \frac{x}{\cos} x$

Cotangent of x equals 0 when the numerator $\cos \left(x\right) = 0$.
This happens when $x = \frac{\pi}{2}$ (there are an infinite amount of values where it becomes 0 but we're just picking the simplest one that comes to mind).

On the other hand, tangent (or negative tangent) of x becomes 0 when the numerator $\sin \left(x\right) = 0$.
This happens when $x = 0$.

Therefore, the two equations are not the same.

You can also see this, by assuming the two equations are equal, and finding a contradiction when making this assumption.

Let's try this.
If the two equations were the same, then their difference should be equal to 0.
We write that statement as such:
$\cot x + \tan x = 0$
(Note: the sign becomes a + because we're doing $\cot x - \left(- \tan x\right)$)

Then we rewrite in terms of sines and cosines:
$\cos \frac{x}{\sin} x + \sin \frac{x}{\cos} x = 0$

which simplifies to:
$\frac{1}{\sin x \cos x} = 0$
(Note: I used the identity ${\sin}^{2} x + {\cos}^{2} x = 1$ to simplify the numerator)
You see now that this is never true for all x, and therefore, that this equality is false.
This means that our assumption ("the two equations are equal") is wrong, which means our two equations are not equal.

You can see it in the following graphs too.
This is $\cot x$:
graph{cot x [-10, 10, -5, 5]}

and this is $- \tan x$:
graph{-tan x [-10, 10, -5, 5]}

Q.E.D.