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Tangent is an odd function tan(-x)=-tan x

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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 $(OA_xA)$ :

$$ \tan x=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{|OA_y|}{|OA_x|} $$

In the triangle $(OA’_xA’)$:

$$ \tan (-x)=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{|OA’_y|}{|OA_x|} $$

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

$$ \forall x\in \mathbb{R},x \neq k\pi + \frac{\pi}{2} , k \in \mathbb{Z}:\quad \tan (-x)=\frac{|OA’_y|}{|OA_x|}=-\frac{|OA_y|}{|OA_x|}=-\tan x $$

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 $x \neq k\pi + \frac{\pi}{2} , k \in \mathbb{Z}$

Using the trigonometric identity for tangent, we can write:

$$\tan(-x) = \frac{\sin(-x)}{\cos(-x)}$$

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.

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  2. Sine is an odd function sin(-x)=-sin x
  3. Tangent is an odd function tan(-x)=-tan x
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