Question

Se cos x= -3/5 e pi/2 menor igual x menor igual pi, o valor da expressão 2tgx/1-tg²x é? 
A) 20/7
B)24/7
C) 1/7
D) 1
E) 0 

Answer 1

Pela relação fundamental, 

$\text{sen}^2~x+\text{cos}^2~x=1$

$\text{sen}^2~x+\left(-\dfrac{3}{5}\right)^2=1$

$\text{sen}^2~x+\dfrac{9}{25}=1$

$\text{sen}^2~x=1-\dfrac{9}{25}$

$\text{sen}^2~x=\dfrac{25-9}{25}$

$\text{sen}^2~x=\dfrac{16}{25}$

$\text{sen}~x=\sqrt{\dfrac{16}{25}}$

Como $\dfrac{\pi}{2}\le x\le\pi$, temos $\text{sen}~x>0$, logo:

$\text{sen}~x=\dfrac{4}{5}$

Com isso, $\text{tg}~x=\dfrac{\text{sen}~x}{\text{cos}~x}=\dfrac{\frac{4}{5}}{-\frac{3}{5}}=-\dfrac{4}{3}$.

$2\cdot\text{tg}~x=2\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{8}{3}$

$1-\text{tg}^2~x=1-\left(-\dfrac{4}{3}\right)^2=1-\dfrac{16}{9}=\dfrac{9-16}{9}=-\dfrac{7}{9}$.

Deste modo:

$\dfrac{2\text{tg}~x}{1-\text{tg}^2~x}=\dfrac{-\frac{8}{3}}{-\frac{7}{9}}=\dfrac{-8}{3}\cdot\dfrac{-9}{7}=\dfrac{72}{21}=\dfrac{24}{7}$

Letra B