Question

me ajudem por favor precIso agora

Answer 1

Para encontrar cós x, lembrar que

${sen}^{2} x + {cos}^{2} x = 1$
Assim,

${ (\frac{15}{17} )}^{2} + {cos}^{2} x = 1 \\ \frac{225}{289} + {cos}^{2} x = 1 \\ {cos}^{2} x = 1 - \frac{225}{289} \\ {cos}^{2} x = \frac{64}{289} \\ cosx = \sqrt{ \frac{64}{289} } \\ cosx = \frac{8}{17}$
Cossec x = 1/ sen x

$cossc \: x = \frac{1}{ \frac{15}{17} } = \frac{17}{15}$
tg x = sen x/ cos x

$tg \: x = \frac{sen \: x}{cos \: x} = \frac{ \frac{15}{17} }{ \frac{8}{17} } = \frac{15}{17} . \frac{17}{8} = \frac{15}{8}$
cotg x = 1/tg x

$cotg \: x = \frac{1}{tg \: x} = \frac{1}{ \frac{15}{8} } = \frac{8}{15}$
sec x = 1/ cos x

$sec \: x = \frac{1}{cos \: x} = \frac{1}{ \frac{8}{17} } = \frac{17}{8}$
Espero ter ajudado