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.
 A: 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 $(1+x)(1+y)$.

A: You've reached here :
$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}
$$
Let's continue it in this way:
$$\large\begin{align}
\Rightarrow & \tan\alpha + \tan\beta+\tan\alpha\tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + \tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + 1+ \tan\beta = 2\\
\Rightarrow &(1 + \tan\alpha)(1 + \tan\beta) = 2\\
\end{align}
$$
A: You were almost there as pointed out by @Daniel in his answer.
Here's another way to do it:
$\beta=\dfrac{\pi}{4}-\alpha\implies (1+\tan\beta)=$  $\left(1+\tan\left(\dfrac{\pi}{4}-\alpha\right)\right)=\left(1+\dfrac{1-\tan\alpha}{1+\tan\alpha}\right)=\left(\dfrac{2}{1+\tan\alpha}\right)$
Thus, $$(1+\tan\beta)=\frac{2}{1+\tan\alpha}$$ $$\implies (1+\tan\alpha)(1+\tan\beta)=2$$
A: 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
