Question

Determine o valor de
\frac{cos x + tgx}{cotg x * sec x} sabendo que sen x = \frac{4}{5} e que pertence ao 1° quadrante

Answer 1

$cos^2x+sen^2x=1\\ cos^2x+ (\frac{4}{5}) ^2=1\\ cos^2x=1- \frac{16}{25} \\ cos^2x = \frac{9}{25} \\\\ cos \ x = \frac{3}{5} ( primeiro\ quadrante\cos x + )$

$\frac{cos\ x +tg\ x}{cotg\ x*sec\ x} = \frac{cos\ x + \frac{sen\ x}{cos\ x} }{ \frac{cos\ x}{sen\ x } \ * \frac{1}{cos\ x}}= \frac{ \frac{cos^2x+sen\ x }{cos \ x} }{ \frac{1}{sen\ x} } = \frac{(cos^2\ x+sen\ x )*sen\ x}{cos \ x} =$

$\frac{(1-sen^2x +sen\ x )*sen\ x}{ cos \ x} } = \frac{(1- \frac{16}{25} + \frac{4}{5})* \frac{4}{5} }{ \frac{3}{5} } = \frac{ \frac{29 }{25}* \frac{4}{5} }{ \frac{3}{5} } = \frac{116}{125} * \frac{5}{3} = \boxed{\frac{116}{75} }$