Answer
$-\frac{3}{5}, \frac{4}{5}, -\frac{3}{4}, -\frac{4}{3}, \frac{5}{4}, -\frac{5}{3}$
Work Step by Step
1. Given $sec\theta=\frac{1}{cos\theta}=\frac{5}{4}$ and $\theta$ in quadrant IV, we have $cos\theta=\frac{4}{5}$. Form a right triangle with sides $\sqrt {5^2-4^2},4,5$ or $3,4,5$, we have:
2. $sin\theta=-\frac{3}{5}$
3. $cos\theta=\frac{4}{5}$
4. $tan\theta=-\frac{3}{4}$
5. $cot\theta=\frac{1}{tan\theta}=-\frac{4}{3}$
6. $sec\theta=\frac{1}{cos\theta}=\frac{5}{4}$
7. $csc\theta=\frac{1}{sin\theta}=-\frac{5}{3}$