Question

me ajudem ?? não estou conseguindo resolver

Answer 1

9)
$cos(x) = \frac{1}{4} \\ cos^{2}(x) = (\frac{1}{4})^{2} \\ 1 - sen^{2}(x) = \frac{1}{16} \\ sen^{2}(x) = 1 - \frac{1}{16} = \frac{16 - 1}{16} = \frac{15}{16} \\ sen(x) = \sqrt{ \frac{15}{16} } \\ sen(x) = \frac{ \sqrt{15} }{4} \\ tg(x) = \frac{sen(x)}{cos(x)} \\ tg(x) = \frac{ \sqrt{15} }{4}.4 = \sqrt{15}$

10)
$sen(x) = \frac{8}{11} \\ sen^{2}(x) = ( \frac{8}{11}) ^{2} \\ 1 - cos^{2}(x) = \frac{64}{121} \\ cos^2(x) = 1 - \frac{64}{121} = \frac{121 - 64}{121} = \frac{57}{121} \\ cos(x) = \frac{ \sqrt{57} }{11}$
$tg(x) = \frac{sen(x)}{cos(x)} \\ tg(x) = \frac{8}{11}. \frac{11. \sqrt{57} }{57} = \frac{8 \sqrt{57} }{57}$
$cotg(x) = \frac{1}{tg(x)} \\ cotg(x) = \frac{57}{8 \sqrt{57} } = \frac{ \sqrt{57} }{8}$

$sec(x) = \frac{1}{cos(x)} \\ sec(x) = \frac{11}{ \sqrt{57} } = \frac{11 \sqrt{57} }{57}$