Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$ 
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.

Attempt:
I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.

Let $$\displaystyle \tan^{-1}\left(\frac{1}{2}\right)=\alpha\Rightarrow \tan \alpha = \frac{1}{2}$$ and $$\displaystyle \tan^{-1}\left(\frac{1}{3}\right)=\beta\Rightarrow \tan \beta = \frac{1}{3}$$
So $$\displaystyle \tan(\alpha+\beta) = \frac{\tan \alpha+\tan \beta}{1-\tan \alpha\cdot \tan \beta} = \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}=1$$
So we get $$\displaystyle \tan(\alpha+\beta)=1=\tan \frac{\pi}{4}\Rightarrow \alpha+\beta = n\pi+\frac{\pi}{4}\;,n\in \mathbb{N}$$
But above $\displaystyle 0<\tan^{-1}\left(\frac{1}{2}\right)<\frac{\pi}{6}$ and $\displaystyle 0<\tan^{-1}\left(\frac{1}{3}\right)<\frac{\pi}{6}$
So we get $\displaystyle \alpha+\beta = \frac{\pi}{4}$
