We prove here that the tangent function tan (-x) = - tan x is odd using the unit circle.

We start with the following configuration:

  • unit circle $\mathcal{C}(O,R=1)$
  • definition of the angle $x$
  • definition of the angle $-x$

Now consider the triangles: $(OA_xA)$ and $(OA’_xA’)$.

Proof that tangent is an odd function tan(-x) = -tan (x)

Take the definition of the tangent of the angles $x$ and $-x$.

In the triangle $(OA_xA)$ :

\[\tan x=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{\vert OA_y\vert }{\vert OA_x\vert }\]

In the triangle $(OA’_xA’)$:

\[\tan (-x)=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{\vert OA'_y\vert }{\vert OA_x\vert }\]

By construction: $\vert OA_y\vert = -\vert OA’_y\vert $, then we have

\[\forall x\in \mathbb{R},x \neq k\pi + \frac{\pi}{2} , k \in \mathbb{Z}:\quad \tan (-x)=\frac{\vert OA'_y\vert }{\vert OA_x\vert }=-\frac{\vert OA_y\vert }{\vert OA_x\vert }=-\tan x\]

Another proof that tangent is an odd function tan(-x) = -tan (x)

The definition of an odd function is that $f(-x) = -f(x)$ for all $x$ in the domain of the function. To prove that the tangent function is odd, we can use this definition and show that $\tan(-x) = -\tan x$ for all $x \neq k\pi + \dfrac{\pi}{2} , k \in \mathbb{Z}$

Using the trigonometric identity for tangent, we can write: \(\tan(-x) = \dfrac{\sin(-x)}{\cos(-x)}\)

Using the fact that sine is odd $\sin(-x) = -\sin x$ and cosine is even $\cos(-x) = \cos x$, we can simplify the above equation to:

\[\tan(-x) = -\left(\frac{\sin x }{\cos x}\right) = -\tan x\]

which means that the tangent function is an odd function.