Question

dado y = sec²(x) - sec(x) .cos sec (x) / 1-cot g (x), e sendo cos (x) = 1/4, pode se afirmar nesse caso que o valor de y é:

16

20

32

24

28

Answer 1

$y = \frac{sec^2(x)-sec(x)*cosec(x)}{1-cotg(x)} \\\\ y= sec(x)* \frac{sec(x)-cosec(x)}{1-cotg(x)} \\\\ y=sec(x)* \frac{ \frac{1}{cos(x)} - \frac{1}{sen(x)} }{1-cotg(x)} \\\\ y=sec(x)* \frac{ \frac{sen(x)-cos(x)}{cos(x)sen(x)} }{1- \frac{cos(x)}{sen(x)} )} \\\\ y=sec(x)* \frac{ \frac{sen(x)-cos(x)}{cos(x)sen(x)} }{ \frac{sen(x)-cos(x)}{sen(x)} } \\\\ y=sec(x)* \frac{sen(x)-cos(x)}{cos(x)sen(x)} * \frac{sen(x)}{sen(x)-cos(x)} \\\\ y=sec(x)* \frac{1}{cos(x)} \\\\ y= (\frac{1}{cos(x)} )^2$

$y= (\frac{1}{ \frac{1}{4} })^2=16$