Step-by-step explanation:
[tex] \tan \theta = \frac{3}{4} \\ \\ \sec \theta = \sqrt{1 + { \tan }^{2} \theta} \\ \hspace{22 pt} = \sqrt{1 + {( \frac{3}{4} )}^{2} } \\ \hspace{22 pt} = \sqrt{1 + {\frac{9}{16} }} \\ \hspace{22 pt} = \sqrt{ {\frac{16 + 9}{16} }} \\ \hspace{22 pt} = \sqrt{ {\frac{25}{16} }} \\ \red{ \boxed{\bold{ \therefore \sec \theta = { {\frac{5}{4} }} }}} \\ \\ \cos \theta = \frac{1}{\sec \theta } = \frac{4}{5} = \frac{a}{b} \\ \\ \implies \: a = 4 \: and \: b = 5[/tex]
