Answer
$cos\theta =\frac{3}{5}$,
$tan\theta =\frac{4}{3}$,
$cot\theta =\frac{3}{4}$,
$sec\theta= \frac{5}{3}$,
$csc\theta= \frac{5}{4}$.
Work Step by Step
Given $sin\theta=\frac{4}{5}$, let $y=4, r=5$. We have $x=\sqrt {r^2-y^2}=3$. As $\theta$ is acute, we know it is in quadrant I, thus the signs of the other functions are also known. We have:
$sin\theta=\frac{y}{r}=\frac{4}{5}$,
$cos\theta=\frac{x}{r}=\frac{3}{5}$,
$tan\theta=\frac{y}{x}=\frac{4}{3}$,
$cot\theta=\frac{x}{y}=\frac{3}{4}$,
$sec\theta=\frac{r}{x}=\frac{5}{3}$,
$csc\theta=\frac{r}{y}=\frac{5}{4}$.
