We have $\tan(x)=\dfrac{\sin(x)}{\cos(x)}$. I was wondering why $\tan(x+{\pi/2})=\tan(x)$?
I wanted to Show
$$\frac{\sin(x+\pi/2)}{\cos(x+\pi/2)}=\frac{\sin(x)}{\cos(x)}\iff\frac{\sin(x+\pi/2)\cos(x)}{\cos(x+\pi/2)\sin(x)}=1\iff\frac{\cos^2(x)}{-\sin^2(x)}=1$$
But this would imply that $\cos^2x+\sin^2x=0$. But this is false
On the other hand we have that $$\tan(x+{\pi/2})=\tan(x)$$
don't we?
On
No.
$$\tan(x+\pi/2)=\frac{\sin(x+\pi/2)}{\cos(x+\pi/2)}$$ $$=\frac{\sin x\cos \pi/2 + \cos x \sin \pi/2}{\cos x \cos \pi/2 - \sin x \sin \pi/2}$$ $$=\frac{\cos x}{-\sin x}$$ $$=-\cot x$$
which is generally not equal to $\tan x$.
The period of $\tan x$ is $\pi$. A similar line of reasoning as above shows that $$\tan (x + \pi/2) = \frac{-\sin x}{-\cos x} = \tan x.$$
On
No, we don't. The real relation is $$\tan\bigl(x+\tfrac\pi 2\bigr)=-\cot x=-\frac 1{\tan x}$$ since $\;\sin\bigl(x+\frac\pi 2)=\cos x$, whereas $\;\cos\bigl(x+\frac\pi 2\bigr)=\color{red}-\sin x$.
What is true is that $\tan x$ has period $\pi$, simply because $$\sin(x+\pi)=-\sin x,\quad\cos(x+\pi)=-\cos x.$$
No, since $\tan\left(\frac\pi4\right)=1$, whereas $\tan\left(\frac\pi2+\frac\pi4\right)=-1$.