Let be 0 an angle in quadrant I such that tqn0

Answer:
[tex]\begin{gathered} \cos \theta=\frac{4}{5} \\ \csc \theta=\frac{5}{3} \end{gathered}[/tex]Step-by-step explanation:
Quadrant 1 on a coordinate plane is the one at the right top corner, therefore the tangent of an angle equals 3/4, would be:
Then, the cos and csc of the same angle would be:
[tex]\begin{gathered} \cos \theta=\frac{\text{ adjacent}}{\text{hypotenuse}} \\ \cos \theta=\frac{4}{\sqrt[]{4^2+3^2}} \\ \cos \theta=\frac{4}{5} \end{gathered}[/tex]For csc theta:
[tex]\begin{gathered} \csc \theta=\frac{1}{\sin \theta} \\ \csc \theta=\frac{hypotenuse}{\text{ opposite}} \\ \csc \theta=\frac{5}{3} \end{gathered}[/tex]