Problem with tangents If $\tan\alpha=3$ and $\tan\beta=2$.  ($\alpha$ and $\beta$ are in first quadrant).

Prove that  $$\frac{\pi}{24}<\alpha-\beta< \frac{\pi}{16}$$

And I get that $$\tan(\alpha-\beta)=\frac{1}{7}$$
But I can't continue
 A: Here is my start on this problem: We need exact answers of trig values through half angle formulas. Using some older trig research, we know that $\sin22.5=\sqrt{\frac{1-0.5\sqrt{2}}{2}}$ and $\cos22.5=\sqrt{\frac{1+0.5\sqrt{2}}{2}}$. This enables us to get an exact value for $tan11.25$ which comes (through it half angle formula $\frac{\sin...}{1+\cos...}$) to be $\frac{\sqrt{1-.5\sqrt{2}}}{\sqrt{2}+\sqrt{1+.5\sqrt{2}}}$ This is the upper value of the inequality (It is a bit more than $1/7$).
Similarly for the lower given angle, I find $\tan7.5=\frac{\sqrt{6}-\sqrt{2}}{4+\sqrt{6}+\sqrt{2}}$. This value is indeed a bit less than 1/7 The problem now is to show numerically that the inequality holds. I am still stuck on that, but there is no trig involved. If someone can add to my solution, that would be great. If it is believed that my approach leads to nothing, then let me know, I will take it off!  
A: This is not yet an answer but hopefully sheds some light.
Looking at the numbers, I think this has something to do with $\pi/8$ because:
$$
\begin{align}
\frac{\pi}{16} &= \frac{1}{2} \frac{\pi}{8} \\
\frac{\pi}{24} &= \frac{1}{3} \frac{\pi}{8}
\end{align}
$$
$\tan(\pi/8)$ can be found using half-angle equations: Trigonometry Angles--Pi/8. We can do the same thing for $\pi/16$ and $\pi/24$. 
A: This is how I understood it. You need first to find the values of alpha and beta. Using the formula alpha=Arctan 3 and beta=Arctan 2 . After getting the values of the two subtract beta to alpha and you will get the answer that  will satisfy what you are proving.
