Question

Se tg (x + y) = 33 e tgx = 3, determine o valor de tg2y.

Answer 1

A tangente da soma de dois arcos:

$\mathrm{tg\,}(x+y)=\dfrac{\mathrm{tg\,}x+\mathrm{tg\,}y}{1-\mathrm{tg\,}x\cdot \mathrm{tg\,}y}\\ \\ \\ 33=\dfrac{3+\mathrm{tg\,}y}{1-3\mathrm{\,tg\,}y}\\ \\ \\ 33\,(1-3\mathrm{\,tg\,}y)=3+\mathrm{tg\,}y\\ \\ 33-99\mathrm{\,tg\,}y=3+\mathrm{tg\,}y\\ \\ 33-3=\mathrm{tg\,}y+99\mathrm{\,tg\,}y\\ \\ 100\mathrm{\,tg\,}y=30\\ \\ \mathrm{\,tg\,}y=\dfrac{30}{100}\\ \\ \mathrm{\,tg\,}y=\dfrac{3}{10}$


Substituindo na relação da tangente do arco duplo:

$\mathrm{tg\,}2y=\dfrac{2\mathrm{\,tg\,}y}{1-\mathrm{tg\,^{2}}y}\\ \\ \\ \mathrm{tg\,}2y=\dfrac{2\cdot \frac{3}{10}}{1-(\frac{3}{10})^{2}}\\ \\ \\ \mathrm{tg\,}2y=\dfrac{\frac{6}{10}}{1-\frac{9}{100}}$


Multiplicando o numerador e o nenominador por $100,$ temos

$\mathrm{tg\,}2y=\dfrac{60}{100-9}\\ \\ \\ \mathrm{tg\,}2y=\dfrac{60}{91}$