Answer
See answers below.
Work Step by Step
1. Given $sin\theta=\frac{3}{7}$ with $\theta$ in quadrant II, we have $y=3, r=7$, thus $x=-\sqrt {7^2-3^2}=-2\sqrt {10}$
2. We have $cos\theta=\frac{x}{r}=-\frac{2\sqrt {10}}{7}$
3. We have $tan\theta=\frac{y}{x}=-\frac{3\sqrt {10}}{20}$
4. We have $cot\theta=\frac{x}{y}=-\frac{2\sqrt {10}}{3}$
5. We have $sec\theta=\frac{1}{cos\theta}=-\frac{7\sqrt {10}}{20}$
6. We have $csc\theta=\frac{1}{sin\theta}=\frac{7}{3}$
