Trigonometria

Giovana Martins escreveu:

Por favor, confira os cálculos.

\\sen(x)=y\to y=arcsen(x)=sen^{-1}(x)\\\\a>0\ \wedge\ sen^{-1}\left ( \frac{a-1}{a+1} \right )\in \mathrm{1^{\circ}\ quadrante}\ \therefore \ cos(x)>0\\\\x=sen^{-1}\left ( \frac{a-1}{a+1} \right )\to sen(x)=\frac{a-1}{a+1}\ \\\\\therefore \ cos(x)=\pm \frac{2\sqrt{a}}{a+1}\ \therefore \ cos(x)=\frac{2\sqrt{a}}{a+1}\ \therefore \ tg(x)=\frac{a-1}{2\sqrt{a}}\\\\y=tg^{-1}\left ( \frac{1}{2\sqrt{a}} \right )\to tg(y)=\frac{1}{2\sqrt{a}}\\\\tg(x+y)=\frac{tg(x)+tg(y)}{1-tg(x)tg(y)}=\frac{2a\sqrt{a}}{3a+1}\\\\\therefore \ \boxed {tg\left [ sen^{-1}\left ( \frac{a-1}{a+1} \right )+tg^{-1}\left ( \frac{1}{2\sqrt{a}} \right ) \right ]=\frac{2a\sqrt{a}}{3a+1}}