Question

Sabendo que a cossec a = -4 e que...

Answer 1

$\displaystyle \underline{\text{Rela{\c c}{\~a}o fundamental da trigonometria e derivadas}}: \\\\ * \text {sen}^2\text x+\text{cos}^2\text x=1\\\\ *1+\text{cotg}^2\text x=\text{cossec}^2\text x \\\\ *\text{tg}^2\text x+ 1=\text{sec}^2$

temos :

$\displaystyle \text{cossec }\alpha = -4 \ \ ; \ \ \frac{3\pi}{2}<\alpha <2\pi \ \ ; \ \text{sec }\alpha =\ ?\\\\ \underline{\text{Usando uma das rela{\c c}{\~o}es}}:\\\\ \text{cossec}^2\alpha =1+\text{cotg}^2\alpha \\\\ (-4)^2=1+\text{cotg}^2\alpha \\\\ \text{cotg}^2\alpha = 16-1 \to \text{cotg}^2\alpha = 15 \\\\ \frac{1}{\text {tg}^2\alpha}=15 \to \boxed{\text{tg}^2\alpha = \frac{1}{15}} \\\\\ \underline{\text{Usando a rela{\c c}{\~a}o da secante}}: \\\\ \text{sec}^2=1+\text{tg}^2\alpha$

$\displaystyle \text{sec}^2\alpha = 1+\frac{1}{15}\\\\\ \text{sec }\alpha =\pm\sqrt{\frac{16}{15}}\\\\ \text{sec }\alpha=\pm\frac{4}{\sqrt{15}} \\\\\\ \text{sec }\alpha = \frac{\pm4\sqrt{15}}{15}\\\\\\ \text{Secante } = \frac{1}{\text{cosseno}} \ \to \ \frac{3\pi}{2}<\alpha <2\pi \ \text{(Quarto quadrante)}$

e o cosseno no 4º quadrante é positivo, então

$\huge\boxed{\text {sec }\alpha = \frac{4\sqrt{5}}{15}}$

letra d