If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$ 
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$

I have had a few ideas about this:
If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$
We also know that $\tan(\alpha +\beta) = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
Then we can write $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
I have tried rearranging $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$ but it has not been helpful.
I also thought if we let $\alpha = \beta$ then I could write $\tan(\alpha+ \alpha) = 1$ 
(does this also mean $\tan(2\alpha) = 1$?)
then: $\tan(\alpha + \alpha) = \dfrac{\tan\alpha + \tan\alpha}{1- \tan\alpha\tan\alpha}$ 
which gives: $1 = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$
Anyway  these are my thoughts so far, any hints would be really appreciated.

You've reached here :
$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}
$$
Let's continue it in this way:
$$\large\begin{align}
\Rightarrow & \tan\alpha + \tan\beta+\tan\alpha\tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + \tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + 1+ \tan\beta = 2\\
\Rightarrow &(1 + \tan\alpha)(1 + \tan\beta) = 2\\
\end{align}
$$
