Question

Dado x=15°, calcular o valor da expressão:​

Answer 1

Vamos lembrar que o arco dobro da tangente é :

$\displaystyle \text{tg(2x)}=\frac{2.\text{tg(x)}}{1-\text{tg}^2(\text x)}$

Temos :

$\displaystyle \text y=\frac{\text{tg(x)}}{1+\text{tg(x)}}+\frac{\text{tg(x)}}{1-\text{tg(x)}} \ \ ; \ \ \text x = 15^\circ \\\\\\ \underline{\text{Vamos arrumar essa express{\~a}o}}:\\\\ \text y = \frac{\text{tg(x)}[1-\text{tg(x)}]+\text{tg(x)}[1+\text{tg(x)}]}{[1+\text{tg(x)}][1-\text{tg(x)}]} \\\\\\ \text y = \frac{\text{tg(x)}-\text{tg}^2(\text x)+\text{tg(x)}+\text{tg}^2(\text x)}{1-\text{tg}^2(\text x)} \\\\\\ \text y = \frac{2\text{tg(x)}}{1-\text{tg}^2(\text x)} \\\\\\ \text{y = tg(2x)}$

Façamos x = 15º

$\displaystyle \text y = \text{tg(2.}15^\circ)\\\\ \text y=\text{tg}(30^\circ) \\\\\ \huge\boxed{\text y = \frac{\sqrt{3}}{3}\ }\checkmark$