Calculate for $(1+\tan 20^\circ)(1+\tan 25^\circ)$. Help me with my works I have no idea what I am doing here, 
I started with $\tan 20^\circ=\tan(45^\circ-25^\circ)=(1-\tan 25^\circ)/(1+\tan 25^\circ)$ 
I am sure the work I have shown so far are ok, but how do you get $1+\tan 20^\circ=2/(1+\tan 25^\circ)$ from that?
 A: It's a consequence of the following  trigonometric identity 
\begin{equation*}
\color{blue}{\left( 1+\tan a\right) \left( 1+\tan b\right) =2+\tan a+\tan b-\dfrac{\tan
a+\tan b}{\tan \left( a+b\right) }.}\tag{1}
\end{equation*}
On  the one hand we rewrite
\begin{equation*}
\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}
\end{equation*}
as
\begin{equation*}
\color {blue}{\tan a\tan b}=1-\frac{\tan a+\tan b}{\tan \left( a+b\right) }.\tag{2}
\end{equation*}
On the other hand setting $x=\tan a$ and $y=\tan b$ in the algebraic identity 
\begin{equation*}
xy=\left( 1+x\right) \left( 1+y\right) -1-x-y
\end{equation*}
yields:
\begin{equation*}
\color {blue}{\tan a\tan b}=\left( 1+\tan a\right) \left( 1+\tan b\right) -1-\tan a-\tan b.\tag{3}
\end{equation*}
If we equate $(3)$ to $(2)$, then we get
\begin{equation*}
\left( 1+\tan a\right)
\left( 1+\tan b\right) -1-\tan a-\tan b=1-\frac{\tan a+\tan b}{\tan \left( a+b\right) },
\end{equation*}
from which $(1)$ follows. For $a=20^{{}^\circ},b=25^{{}^\circ}$ we obtain
\begin{eqnarray*}
\left( 1+\tan 20{{}^\circ}\right) \left( 1+\tan 25{{}^\circ}\right)  
&=&2+\tan 20{{}^\circ}+\tan 25{{}^\circ}-\frac{\tan 20{{}^\circ}+\tan 25{{}^\circ}}{\tan \left( 45{{}^\circ}\right) } \\
&=&2+\tan 20{{}^\circ}+\tan 25{{}^\circ}-\frac{\tan 20{{}^\circ}+\tan 25{{}^\circ}
}{1} \\
&=&2.\tag {4}
\end{eqnarray*}
A: $1 = \tan(45^\circ) = \tan(20^\circ+25^\circ) = \frac{ \tan 20^\circ + \tan 25^\circ}{1- \tan 25^\circ \tan 20^\circ} $
so
$2 = 1+ \tan 20^\circ + \tan 25^\circ + \tan 20^\circ \tan 25^\circ = (1+ \tan 20^\circ)(1+ \tan 25^\circ)$. 
