Question

sabendo que tg(x) = 1/5 e tg(y)=1/10, calculando o valor de tg(x+y) obtemos:

(na "/" lê-se "sobre")

Answer 1

Identidade da tangente da soma de dois arcos:

$\boxed{\begin{array}{c}\mathrm{tg}(x+y)=\dfrac{\mathrm{tg\,}x+\mathrm{tg\,}y}{1-\mathrm{tg\,}x\cdot \mathrm{tg\,}y} \end{array}}~~~~~~~~\text{com }\mathrm{tg\,}x\cdot \mathrm{tg\,}y\ne 1$

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Para esta questão, temos

$\mathrm{tg\,}x=\dfrac{1}{5}$ e $\mathrm{tg\,}y=\dfrac{1}{10}$


Portanto,

$\mathrm{tg}(x+y)=\dfrac{\frac{1}{5}+\frac{1}{10}}{1-\frac{1}{5}\cdot \frac{1}{10}}\\\\\\ \mathrm{tg}(x+y)=\dfrac{\frac{2}{10}+\frac{1}{10}}{1-\frac{1}{50}}\\\\\\ \mathrm{tg}(x+y)=\dfrac{\frac{3}{10}}{\frac{50}{50}-\frac{1}{50}}\\\\\\ \mathrm{tg}(x+y)=\dfrac{\left(\frac{3}{10} \right )}{\left(\frac{49}{50} \right )}\\\\\\ \mathrm{tg}(x+y)=\dfrac{3}{10}\cdot \dfrac{50}{49}\\\\\\ \mathrm{tg}(x+y)=\dfrac{3}{\diagup\!\!\!\!\! 10}\cdot \dfrac{\diagup\!\!\!\!\! 10\cdot 5}{49}\\\\\\ \boxed{\begin{array}{c}\mathrm{tg}(x+y)=\dfrac{15}{49} \end{array}}$