Question: > 
> What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?
> 
> 
> 

I'm very new to the mathjax formatting and the site in general but when I work the problem out it seems as though the answer should be $-1$.

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1+\left(-1\cdot\tan\left(\theta\right)\right)}$$

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1-\tan\left(\theta\right)}$$

Answer: 
In general -- for any functions, not just for trig functions --
if $g(\theta) = f(\theta - k)$ for all $\theta,$ where $k$ is a constant,
then the graph of $g$ is the graph of $f$ shifted $k$ units to the right.

If $h(\theta) = g(-\theta),$ then the graph of $h$ is the mirror image of the graph of $h$, flipped left to right and right to left across the $y$ axis.

So the graph of $\tan\left(\frac{7\pi}{4} - \theta\right)$ is the mirror image of
the graph of $\tan\left(\theta - \frac{7\pi}{4}\right),$ which in turn is
the graph of $\tan\left(\theta \right)$ shifted $\frac{7\pi}{4}$ units to the right.

This does not tell you the simplest way to write the function as a formula
(in fact you can't get much simpler than it already is), but it should help you to avoid stumbling into wrong formulas. Whatever formula you come up with has to have a graph that looks a lot like the graph of $\tan(\theta)$, just mirror-imaged and moved somewhat sideways. It is certainly not a flat line at height $-1.$
$$x=\tan\left(\frac{7\pi}{4}-\theta\right)
=-\tan\left(θ + \frac{\pi}{4}\right)$$

$$0 \le\theta\le\pi\implies -\infty\le x \le 1$$

Note: $\space
\theta\in\{0,\space\frac{\pi}{2}\}\implies 
x=-1\space\land\space
\theta=\pi\implies x=1\space$

and as
$${\theta\rightarrow\frac{\pi}{4}}\implies x\rightarrow - \infty
$$