Question

se tg$\alpha$ = 1/3, calcular tg2$\alpha$

Answer 1

$tg 2 \alpha = tg( \alpha + \alpha )$

$tg( \alpha + \beta ) = (tg \alpha +tg \beta)/(1-tg \alpha *tg \beta )$
$tg ( \alpha + \alpha) = (tg \alpha +tg \alpha )/(1-tg \alpha *tg \alpha )$
$tg 2 \alpha = 2tg \alpha / (1 - tg^{2} \alpha )$
$tg 2 \alpha = 2*[1 / 3] / (1-[1/3]^{2})$
$tg 2 \alpha = (2 / 3) / (1 - [1/9])$
$tg 2 \alpha = (2 / 3) / ([9 / 9] - [1 / 9])$
$tg2 \alpha = (2 / 3) / ([9 - 1] / 9)$
$tg2 \alpha = (2 / 3) / (8 / 9)$
$tg2 \alpha = (2 / 3) * (9 / 8)$
$tg2 \alpha = (1/1)*(3/4)$
$tg2 \alpha =3/4$