tan 100° + tan 125° + tan 100° tan 125° is equal to

$\begin{array}{l}\tan \left({100}^{\circ }+{125}^{\circ }\right)=\frac{\tan {100}^{\circ }+\tan {125}^{\circ }}{1-\tan {100}^{\circ }\tan {125}^{\circ }}\\ \therefore \tan {225}^{\circ }=\frac{\tan {100}^{\circ }+\tan {125}^{\circ }}{1-\tan {100}^{\circ }\tan {125}^{\circ }}\\ \mathrm{i}.\mathrm{e}., 1=\frac{\tan {100}^{\circ }+\tan {125}^{\circ }}{1-\tan {100}^{\circ }\tan {125}^{\circ }}\end{array}$
i.e., tan100° + tan 125° + tan 100° tan 125° = 1