Question

Prove the following :
$\tan {18}^o+\tan {27}^o+\tan {18}^o.\tan {27}^o=?$

Answer 1

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$⟹\tan ({18}^o+{27}^o)=\frac{\tan {18}^o+\tan {27}^o}{1-\tan {18}^o\tan {27}^o}$
$⟹\tan \left({45}^o\right)=\frac{\tan {18}^o+\tan {27}^o}{1-\tan {18}^o\tan {27}^o}$
$⟹1=\frac{\tan {18}^o+\tan {27}^o}{1-\tan {18}^o\tan {27}^o}$
$⟹1-\tan {18}^o\tan {27}^o=\tan {18}^o+\tan {27}^o$
$⟹1=\tan {18}^o+\tan {27}^o+\tan {18}^o\tan {27}^o$
Hence, proved.