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.

from where OP left his step:$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$$
$$\implies{1- \tan\alpha\tan\beta}=\tan\alpha + \tan\beta$$
$$\implies \tan\alpha + \tan\beta+\tan\alpha\tan\beta=1$$
add 1 to both sides
$$\implies\tan\alpha + \tan\beta+\tan\alpha\tan\beta+1=2$$
$$\implies1+\tan\alpha + \tan\beta+\tan\alpha\tan\beta=2$$
factor the above equation
$$\implies1(1+\tan\alpha) + \tan\beta(1+\tan\alpha)=2$$
$$\implies(1+\tan\alpha)(1+\tan\beta)=2$$
Hence proven
