Still incorrect problem. Without a calculator you know that both $\sin$ values are in the interval $(\frac{\sqrt{2}}{2},1)$, so the expression is at least $2\frac{\sqrt{2}}{2} + (\frac{\sqrt{2}}{2})^2 \approx 1.4 + 0.5 = 1.9\dots$. Therefore the best guess option would be 2. But this far away from the true value $2.610667\dots$
$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{{2\cos\pars{40^{\circ}} - \cos\pars{20}^{\circ} \over \sin\pars{20}^{\circ}}:
\ {\large ?}}$
\begin{align}
&\color{#66f}{\large{2\cos\pars{40^{\circ}} - \cos\pars{20}^{\circ} \over \sin\pars{20}^{\circ}}}
={2\cos\pars{40^{\circ}}\cos\pars{60^{\circ}} - \cos\pars{20}^{\circ}\cos\pars{60^{\circ}}
\over \sin\pars{20}^{\circ}\cos\pars{60^{\circ}}}
\\[3mm]&={\cos\pars{40^{\circ}}
-\
\overbrace{\bracks{\cos\pars{60^{\circ}}\cos\pars{20}^{\circ} + \sin\pars{60^{\circ}}\sin\pars{20^{\circ}}}}^{\ds{=\ \cos\pars{40^{\circ}}}}\ +\ \sin\pars{60^{\circ}}\sin\pars{20^{\circ}}
\over \sin\pars{20}^{\circ}\cos\pars{60^{\circ}}}
\\[3mm]&=\tan\pars{60^{\circ}} = \color{#66f}{\large\root{3}} \approx 1.7321
\end{align}
Best Answer
Hint: $\tan(\alpha + \beta)=\dfrac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ and $\tan(225^\circ)=1$.