Question

$\tan {18}^o+\tan {27}^o+\tan {18}^o\cdot \tan {27}^o=1$

Answer 1

$\tan 18+\tan 27+\tan 18\tan 27=1$

We know that $\tan 45=1$

$\tan (18+27)=1\frac{\tan 18+\tan 27}{1-\tan 18\tan 27}=1\tan 18+\tan 27=1-\tan 18\tan 27\therefore \tan 18+\tan 27+\tan 18\tan 27=1$