Question

Questão) Seja 2. arc tg $\frac{1}{3}$ + arc tg $\frac{1}{7}$ = θ. O valor de tg θ é igual a:


a) 1

b) 1/2

c) 1/3

d) 1/14

e) 1/21

Answer 1

Explicação passo-a-passo:

• $\sf tg~(2x)=\dfrac{2\cdot tg~(x)}{1-tg^2~(x)}$

$\sf tg~(2x)=\dfrac{2\cdot\frac{1}{3}}{1-\Big(\frac{1}{3}\Big)^2}$

$\sf tg~(2x)=\dfrac{\frac{2}{3}}{1-\frac{1}{9}}$

$\sf tg~(2x)=\dfrac{\frac{2}{3}}{\frac{9-1}{9}}$

$\sf tg~(2x)=\dfrac{\frac{2}{3}}{\frac{8}{9}}$

$\sf tg~(2x)=\dfrac{2}{3}\cdot\dfrac{9}{8}$

$\sf tg~(2x)=\dfrac{18}{24}$

$\sf tg~(2x)=\dfrac{3}{4}$

• $\sf tg~(x+y)=\dfrac{tg~(x)+tg~(y)}{1-tg~(x)\cdot tg~(y)}$

$\sf tg~\theta=\dfrac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\cdot\frac{1}{7}}$

$\sf tg~\theta=\dfrac{\frac{21+4}{28}}{1-\frac{3}{28}}$

$\sf tg~\theta=\dfrac{\frac{25}{28}}{\frac{28-3}{28}}$

$\sf tg~\theta=\dfrac{\frac{25}{28}}{\frac{25}{28}}$

$\sf tg~\theta=\dfrac{25}{28}\cdot\dfrac{28}{25}$

$\sf \red{tg~\theta=1}$

Letra A