Does $-\tan(x)=\tan(-x)$ for all $x$?
Just to clarify, does $-\tan(x)=\tan(-x)$ for all $x$?
$\endgroup$ 44 Answers
$\begingroup$Yes.
Often, when I have questions about things like this, I rewrite everything in terms of $\sin$ and $\cos$, and simplify it using what I know about those.
This also easily follows from the unit circle description of trig functions.
$\endgroup$ $\begingroup$$$\tan (x) = \frac{\sin (x)}{\cos (x)}$$ $$\tan (-x) = \frac{\sin (-x)}{\cos (-x)}$$ $$\cos (-x) = \cos (x)$$ $$\sin (-x) = -\sin (x)$$ $$\tan (-x) = \frac{-\sin (x)}{\cos (x)}$$ $$= -\frac{\sin (x)}{\cos (x)}$$ $$= -\tan (x)$$
$\endgroup$ $\begingroup$Recall that cosine is an even and sine an odd function. This means that $\cos (-x) = \cos x$ and $\sin (-x) = -\sin x$, a fact which you can easily verify by checking their respective graphs. It follows from the basic properties of real numbers that the quotients $\sin x / \cos x$ and $\cos x / \sin x$ are odd.
$\endgroup$ $\begingroup$$$\text{Putting }A=0,\text{ in }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\cdot\tan B},$$
$$\tan(-B)=\frac{\tan 0-\tan B}{1+\tan 0\cdot\tan B}=-\tan B\text{ as }\tan0=0$$
$\endgroup$
