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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.

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  • $\begingroup$ You should give the results of your own attempts (and explain why they're unsatisfactory), so that people don't duplicate your effort or give you solutions you don't want. (You could/should leave out the unnecessary $a$ factor, as you did in the title.) $\endgroup$ – Blue Sep 23 '18 at 8:11
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    $\begingroup$ @Blue Thanks! I edited out the $a$ and added my own attempts. $\endgroup$ – Ian H. Sep 23 '18 at 8:21
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    $\begingroup$ @IanH. If you expand it further you will get $$\frac{\sin(\alpha)}{\cos(\beta)\cos(\alpha+\beta)}$$ $\endgroup$ – paulplusx Sep 23 '18 at 8:32
  • $\begingroup$ @paulplusx May I ask how you got to this result? $\endgroup$ – Ian H. Sep 23 '18 at 8:33
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    $\begingroup$ @Blue No need to delete. Your way is much shorter and faster. I merely showed OP a way to continue from where he/she left. Edit: Sadly, you have already deleted by the time I posted this comment :-( if you like you may repost it :-) $\endgroup$ – paulplusx Sep 23 '18 at 8:53
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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.

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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.

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$$\tan(\alpha+\beta)-\tan\beta=\dfrac{\sin(\alpha+\beta-\beta)}{\cos(\alpha+\beta)\cos\beta}=?$$

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