Question

1. If sin x =3/5 and cos y =12/13, evaluate:

i)sec2 x

ii)tan x + tan y

Answer 1

$\sin(x) = \frac{3}{5} = \frac{perpendicular}{hypotenuse} \\ \cos(x) = \frac{base}{hypotenuse} \\ {h}^{2} = {b}^{2} + {p}^{2} \\ 25 = {b}^{2} + 9 \\ {b}^{2} = 25 - 9 = 16 \\ base = 4 \\ \cos(x) = \frac{4}{5} \\ \sec(x) = \frac{1}{ \cos(x) } \\ = \frac{1}{ \frac{4}{5} } = \frac{5}{4} \\ \sec(2x) = \frac{5}{2} \\ \cos(y) = \frac{12}{13} \\ sin(y ) = \frac{5}{13} \\ \tan(x) + \tan(y) = \frac{ \sin( x ) }{ \cos(x) } + \frac{ \sin(y) }{ \cos(y) } \\ = \frac{ \frac{3}{5} }{ \frac{4}{5} } + \frac{ \frac{5}{13} }{ \frac{12}{13} } \\ = \frac{3}{4} + \frac{5}{12} \\ = \frac{9 + 5}{12} \\ = \frac{14}{12} = \frac{7}{6}$

Answer 2

17. number is your answer ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤