Yes, this would be nice if the "co"s matched up! However, it is indeed all geometry related. No calculus necessary, all you need is to look at the unit circle. Check out this webpage.
One thing to keep in mind is that a "secant line" is just a line that "cuts" through a figure. Secant has a root word in Latin (secare, I believe) which means "to cut". So, the secant should somehow involve "cutting" something, and the webpage above shows you the geometric idea behind the secant. It's the blue line that shows up.
Hopefully that makes it easier to forgive the co-confusion it invariably causes.
It depends on how you look at it I guess, but:
$$\cot(x) = \frac{1}{\tan(x)}$$
$$\csc(x) = \frac{1}{\sin(x)}$$
$$\sec(x) = \frac{1}{\cos(x)}$$
So the three "extra" functions your friend told you about are just derived from the three you know. But if that's the rule, then two of the ones you know,
$$\cos(x) = \sin\left(\frac{\pi}{2} - x\right)$$
$$\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\sin{x}}{\sin\left(\frac{\pi}{2} - x\right)}$$
are also just derived functions. Hence we would say there is only one trigonometric function, for example $\sin{x}$.
(As others have mentioned, this statement works even counting hyperbolic functions, because of properties like $\cosh(x) = \cos(ix)$ and so on, or using $e^{i\theta} = \cos{\theta} + i\sin{\theta}$. But since you don't appear to be at this level of math yet, I won't go into detail about that.)
Bottom line: We only need one trigonometric function, but for practical reasons, there are more.
Best Answer
Their graphs are often useful... though maybe not in rectangular coordinates. Plot $r=\sec(\theta) $ or $ r=\csc(\theta)$ in polar coordinates - you get a straight line a distance $1$ from the origin. Consequently, they are useful when you want to solve "straight line" problems in polar coordinates.
In general, polar equations are nice in physics problems because macroscopic force laws often are dependent only on the distance between two particles. You might end up using secant if you are calculating electrical interactions between a point charge (the origin) and an infinite straight charged wire at a distance $d$, parametrized by $r=d\sec(\theta)$. The contribution of any point on the wire to the electric field will depend on its distance $d\sec(\theta)$ from the origin.
Another example: if you are standing a distance $d$ from a long straight road watching a car go along, and you want to find the rate $d\theta/dt$ you have to turn your head to watch the car at a certain point, I believe you will find yourself differentiating $\sec$ or $\csc$ if you solve it as a related rates problem.