Answer
$sec~\theta=\frac{\sqrt {13}}{2}$
$cot~\theta=\frac{2}{3}$
$csc~\theta=\frac{\sqrt {13}}{3}$
$cos~\theta=\frac{2\sqrt {13}}{13}$
$sin~\theta=\frac{9\sqrt {13}}{13}$
Work Step by Step
$acute~angle:~(sin~\theta\gt0$ and $cos~\theta\gt0)$
$sec^2\theta=1+tan^2\theta=1+(\frac{3}{2})^2=\frac{13}{4}$
$sec~\theta=\frac{\sqrt {13}}{2}$
$cot~\theta=\frac{1}{tan~\theta}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$
$csc^2\theta=cot^2\theta+1=(\frac{2}{3})^2+1=\frac{13}{9}$
$csc~\theta=\frac{\sqrt {13}}{3}$
$cos~\theta=\frac{1}{sec~\theta}=\frac{2}{\sqrt {13}}=\frac{2\sqrt {13}}{13}$
$sin~\theta=\frac{1}{csc~\theta}=\frac{9}{\sqrt {13}}=\frac{9\sqrt {13}}{13}$
