Further simplify $\tan(\alpha+\beta)-\tan(\beta)$ I came up with the formula 
\begin{align*}
\tan(\alpha+\beta)-\tan(\beta)
\end{align*}
but I keep wondering, whether it's possible to further simplify this, into for example only using the $\tan$ once. I tried using the addition theorems for trigonometry, but these just seem to complicate them further.
I already tried something along this:
\begin{align*}
r & =\tan(\alpha+\beta)-tan(\beta) \\
& = \frac{\tan\alpha + \tan \beta}{1 - \tan\alpha\tan\beta}-\tan\beta \\
& = \frac{\tan\alpha + \tan \beta}{1 - \tan\alpha\tan\beta}-\frac{(1-\tan\alpha\tan\beta)(\tan\beta)}{1 - \tan\alpha\tan\beta} \\
& = \frac{(\tan\alpha+\tan\beta)-(\tan\beta-\tan\alpha\tan^2\beta)}{1-\tan\alpha\tan\beta} \\
& = \frac{\tan\alpha+\tan\alpha\tan^2\beta}{1-\tan\alpha\tan\beta}
\end{align*}
but at this point I am pretty stuck on what to try next.
 A: Continuing from where you left:
$$\frac{\tan\alpha+\tan\alpha\tan^2\beta}{1-\tan\alpha\tan\beta}=\frac{\tan\alpha(1+\tan^2\beta)}{1-\tan\alpha\tan\beta}$$
Now $\displaystyle 1+\tan^2\beta=\sec^2\beta=\frac{1}{\cos^2\beta}$, so we have:
$$\begin{align*}\frac{\tan\alpha(1+\tan^2\beta)}{1-\tan\alpha\tan\beta}&=\frac{\frac{\sin\alpha}{\cos\alpha\cos^2\beta}}{1-\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}\\&=\frac{\sin\alpha}{cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)}\end{align*}$$
Now we know, $\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta)$
So you have:
$$\frac{\sin(\alpha)}{\cos(\beta)\cos(\alpha+\beta)}$$
Edit: I just showed you how to simply from where you left. Instead of doing that, directly convert your identities to $\sin$ and $\cos$ as @Blue suggested. It would be faster and would require very less effort compared to this roundabout way.
A: The expression
$$\tan(\alpha+\beta)-\tan(\beta)$$ goes to infinity for $\alpha+\beta=\dfrac\pi2+m\pi$ and $\beta=\dfrac\pi2+n\pi$.
Assuming it can be expressed as a fraction, the denominator must have roots at these values, i.e. have the factors $\cos(\alpha+\beta)\cos(\beta)$, or $\sin(\alpha)\cos(\beta)$.
The numerator is obviously not a constant, so that it doesn't seem possible to simplify, i.e. to express the same quantity with less than two trigonometric functions.
A: $$\tan(\alpha+\beta)-\tan\beta=\dfrac{\sin(\alpha+\beta-\beta)}{\cos(\alpha+\beta)\cos\beta}=?$$
