<s>[INST] 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.
 [/INST] You were almost there:
$$\begin{align}
1 &= \frac{\tan \alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\\
1 - \tan\alpha\tan\beta &= \tan\alpha + \tan\beta\\
2 &= 1 + \tan\alpha + \tan\beta + \tan\alpha\tan\beta\\
2 &= (1+\tan\alpha)(1+\tan\beta)
\end{align}$$
where each equation is equivalent to the preceding/following, and the two can be transformed into each other by a simple step.


*

*First multiply with the denominator (that's $\neq 0$),

*then add $1 + \tan\alpha\tan\beta$ to both sides,

*then write $1 + x + y + xy$ as the product $(1+x)(1+y)$.

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