Question

Prove that tan(nπ+alpha) = tan alpha

Answer 1

$\tan(n\pi + \alpha ) = \frac{ \tan(n\pi) + \tan( \alpha ) }{ 1 - \tan(n\pi) \tan( \alpha ) }$


two cases : (1) n =integer
that means tan(nα)=0
so RHS reduces to

$= \frac{0 + \tan( \alpha ) }{1 - 0} \\ = \tan( \alpha )$


case 2 : n = rational (fraction)


NOT VALID