Question

Prove que (sen alfa + cos alfa)^2 / 1 - sen^2 alfa. - (sen alfa - cos alfa / cos alfa)^2. = 4tg alfa

Answer 1

$\dfrac{(sen \ x+cos\ x)^2}{1-sen^2 x} - \left(\dfrac{sen \ x - cos \ x}{cos \ x}\right)^2 \\ \\ \dfrac{sen^2x + 2sen \ x \ cos \ x+ cos^2x}{cos^2 x} - \left(\dfrac{sen \ x - cos \ x}{cos \ x}\right)^2 \\ \\ \dfrac{1 + 2sen \ x \ cos \ x}{cos^2 x} - \left(\dfrac{sen \ x - cos \ x}{cos \ x}\right)^2 \\ \\ \dfrac{1 + 2sen \ x \ cos \ x}{cos^2 x} - \dfrac{sen^2 x - 2sen \ x cos \ x+ cos^2 \ x}{cos^2x} \\ \\ \dfrac{1 + 2sen \ x \ cos \ x}{cos^2 x} - \dfrac{1 - 2sen \ x cos \ x}{cos^2x}$
$\dfrac{1 + 2sen \ x \ cos \ x - 1 + 2sen \ x \ cos \ x}{cos^2x} \\ \\ \dfrac{ 4sen \ x \ cos \ x}{cos^2x} \\ \\ \dfrac{ 4sen \ x }{cosx} \\ \\ 4 \ tg \ x$