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.
According to the written formula,
$\tan \theta = \frac {\text{opposite side}}{\text{adjacent side}} = \frac b{\text{horizontal radius}} = \frac b1 = b$
The secant can be tackled similarly.
Best Answer
$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{|BE|}{|BC|} = |BE|$
since $|BC|=1$. Similarly
$\cot(\theta) = \frac{\text{adjacent}}{\text{opposite}} = \frac{|AB|}{|BC|} = |AB|$
The placing of the functions in the diagram is purely to do with finding a triangle in which one angle is $\theta$ and the denominator of the relevant ratio is 1 because it is a radius of the circle.