Why does tan(t) touch the unit circle at (1,0)? 
I can't get my head around this, any help would be very much appreciated.
Thanks
EDIT: t is an angle, where 0 < t  < 90, angle t is in degrees
EDIT: Added a picture I lifted from google
 A: What the picture shows is that the right triangle $OSQ$ is similar to the right triangle $OPR$. In particular, $$\tan(h)=\frac{\sin(h)}{\cos(h)}$$ by definition, and trivially, $$\tan(h)=\frac{\tan(h)}1.$$ The length of $OS$ is $\cos(h)$ ahd the length of $SQ$ is $\sin(h)$ by definition. Since it's a unit circle, then the length of $OP$ is $1$, so by similarity, the length of $PR$ is therefore $\tan(h)$.
A: I think what you're trying to express is that the length of the vertical segment joining $P=(1,0)$ to the point labelled $R$ in the diagram is the value of $\tan h$ where $h$ is the angle in the diagram. When $R$ is above the $x$ axis (as in the diagram), this is true by using the definition of $\tan h$ as "opposite over adjacent", because in this picture the "adjacent" has length $1$, being the segment from $O=(0,0)$ to $P=(1,0).$
The same thing works if $R$ happens to be below the $x$ axis, provided we interpret the segment joining $P$ to $R$ as having negative length (i.e. "signed length" is negative) because it extends downward from $P$ Think of the vertical line through $P$ as a real axis, labelled with numbers going positive in the upward direction.
A: Let us find the intersection of the Unit Circle is $x^2+y^2=1$ with the line $y=\tan t$
So, $x^2=1-y^2=1-\tan^2t\implies x^2+\tan^2t-1=0$
This is a Quadratic equation in $x$
For tangency, the two roots must be equal i.e., the  discriminant $(0)^2-4\cdot1\cdot(\tan^2t-1)$ must be $0$
$\implies \tan^2t=1\implies \tan t=1$ as $0^\circ<t<90^\circ$
$\implies x=0, y=1$
So, $y=\tan t$ is tangent of the Unit Circle  at $(0,1)$
