Question: > 
> What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?
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> 
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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: 
The answer would be $-1$ only if numerator was the negative of denominator. But it isn't.
$$x=\tan\left(\frac{7\pi}{4}-\theta\right)
=-\tan\left(θ + \frac{\pi}{4}\right)$$

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

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

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