The value of $\tan {40}^0 + \tan {11}^0 + \tan {20}^0 - \tan {56}^0+\tan {56}^0 \tan {11}^0 + \sqrt{3} \tan {40}^0 \tan {20}^0 =$ 

$$\begin{gathered} \tan {60}^{\circ }=\tan ({40}^{\circ }+{20}^{\circ }) \\ \sqrt{3}=\frac{\tan {40}^{\circ }+\tan {20}^{\circ }}{1-\tan {40}^{\circ } \tan {20}^{\circ }} \\ \sqrt{3}-\sqrt{3} \tan {40}^{\circ } \tan {20}^{\circ }=\tan {40}^{\circ }+\tan 2^{\circ } \\ \tan {40}^{\circ }+\tan {20}^{\circ }+\sqrt{3}\tan {40}^{\circ }\tan {20}^{\circ }=\sqrt{3} \end{gathered}$$
$$\begin{gathered} \tan {45}^{\circ }=\tan ({56}^{\circ }-{11}^{\circ }) \\ 1=\frac{\tan {56}^o-\tan {11}^o}{1+\tan {56}^o\tan {11}^o} \\ \tan {56}^o\tan {11}^o-\tan {56}^o+\tan {11}^o=-1 \end{gathered}$$
$\tan {40}^0 + \tan {11}^0 + \tan {20}^0 - \tan {56}^0+\tan {56}^0 \tan {11}^0 + \sqrt{3} \tan {40}^0 \tan {20}^0 =\sqrt{3}-1$