I have no idea where to start.
tan(pi/2-x)=cotx
sec(pi/2-x)=cscx
csc(pi/2-x)=secx
For the first one, I did the tan(a-b)=tan(a)+tan(b)/1+tan(a)tan(b), but I got -tan(x)/1.
So confused!
Tangent isn't defined at $\displaystyle \frac{\pi}{2}$. The identity doesn't seem possible in my opinion. Secant also is not defined at $\displaystyle \frac{\pi}{2}$.
Edit: I was wrong, the identities DO seem to be true. I was mistaken because when I went to prove it with the tangent of a difference, the $\displaystyle tan(\frac{\pi}{2})$ is undefined. What you need to do, is write the tangent as $\displaystyle \frac{sin(\frac{\pi}{2}-x)}{cos(\frac{\pi}{2}-x)}$. So when you expand, all of the functions are defined on $\displaystyle \frac{\pi}{2}$
For csc and the sec part
I did csc(pi/2-x)= 1/sin(pi/2-x) then I changed it to
1/ sin(pi/2)cos(x)-cos(pi/2)sin(x)= 1/cosx = sec(x)
I did it almost the same, but I didn't put pi/2-x on the numerator and denominator like how you did for tan. Did I do it wrong?
Yes, just write them in terms of either sine, or cosine. Sine and cosine are defined for all real numbers, so you can't go wrong. When you tried to expand the tangent of a difference, you had terms with $\displaystyle tan(\frac{\pi}{2})$. The problem is that this number doesn't exists, this is why the graph of tangent has an asymptote at $\displaystyle \frac{\pi}{2}$.
To do prove the cofunction theorem for cosecant, just expand:
$\displaystyle csc(\frac{\pi}{2}-x)=\frac{1}{sin(\frac{\pi}{2}-x)}$
If you expand this, you should get $\displaystyle sec(x)$.