Question

Sendo seno 2B= 1/3 determine TgB + CotgB

Answer 1

Seno do arco duplo:

$sen(2B)=2senB.cosB=\frac{1}{3}\\ \\ \boxed{senB.cosB=\frac{1}{6}}$

Agora:

$tgB+cotgB=\frac{senB}{cosB}+\frac{cosB}{senB}=\frac{sen^2B+cos^2B}{senB.cosB}=\\ \\ =\frac{1}{\frac{1}{6}}=6$

Answer 2

Lembre-se que:

$\text{sen}~(2x)=2\cdot\text{sen}~x\cdot\text{cos}~x$

Pelo enunciado, $\text{sen}~2B=\dfrac{1}{3}$.

Assim, $2\cdot\text{sen}~B\cdot\text{cos}~B=\dfrac{1}{3}$, donde, $\text{sen}~B\cdot\text{cos}~B=\dfrac{1}{6}$

Queremos determinar $\text{tg}~B+\text{cotg}~B$.

Lembre-se que, $\text{tg}~B=\dfrac{\text{sen}~B}{\text{cos}~B}$ e $\text{cotg}~B=\dfrac{\text{cos}~B}{\text{sen}~B}$.

Assim, $\text{tg}~B+\text{cotg}~B=\dfrac{\text{sen}~B}{\text{cos}~B}+\dfrac{\text{cos}~B}{\text{sen}~B}=\dfrac{\text{sen}^2~B+\text{cos}^2~B}{\text{sen}~B\cdot\text{cos}~B}$.

Pela relação fundamental, $\text{sen}^2~B+\text{cos}^2~B=1$. Logo:

$\text{tg}~B+\text{cotg}~B=\dfrac{1}{\frac{1}{6}}=6$.