Calculate for $(1+\tan 20^\circ)(1+\tan 25^\circ)$. Help me with my works

Question

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?

Answer 1

$1 = \tan(45^\circ) = \tan(20^\circ+25^\circ) = \frac{ \tan 20^\circ + \tan 25^\circ}{1- \tan 25^\circ \tan 20^\circ} $

so

$2 = 1+ \tan 20^\circ + \tan 25^\circ + \tan 20^\circ \tan 25^\circ = (1+ \tan 20^\circ)(1+ \tan 25^\circ)$.

Answer 2

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*}