9.1 Reference Triangles & Inverse Trig Functions

$$\sqrt[]{2}$$

$$\frac{\sqrt[]{2}}{2}$$

1

$$\frac{\sqrt[]{2}}{23}$$

$$\frac{13\sqrt[]{5}}{22}$$

$$\frac{22\sqrt[]{5}}{65}$$

$$\frac{13\sqrt[]{5}}{19}$$

$$\frac{19\sqrt[]{5}}{65}$$

$$\frac{22}{19}$$

$$\frac{22\sqrt[]{5}}{65}$$

$$\frac{19}{22}$$

$$\frac{19\sqrt[]{5}}{65}$$

$$\sin\theta=\cos\theta$$

$$\tan\theta=\cot\theta$$

$$\sin\theta=\csc\theta$$

$$\cos\theta=\sec\theta$$

$$\tan\theta=-\frac{7}{12}$$

$$\tan\theta=-\frac{12}{7}$$

$$\tan\theta=-\frac{12\sqrt[]{193}}{193}$$

$$\tan\theta\pm-\frac{7\sqrt[]{193}}{7}$$

$$\sin\theta=\frac{\sqrt[]{5}}{3}$$

$$\sin\theta=\frac{3\sqrt[]{5}}{5}$$

$$\sin\ \theta=\frac{\sqrt[]{5}}{2}$$

$$\sin\theta=\frac{2\sqrt[]{5}}{5}$$

$$\csc\theta=\frac{5}{3}$$

$$\csc\theta=\frac{-4}{5}$$

$$\csc\theta=\frac{\sqrt[]{7}}{3}$$

$$\csc\theta=\frac{4\sqrt[]{7}}{7}$$

$$\cos\ \theta>0$$

$$\tan\theta>0$$

$$\csc\theta>0$$

$$\theta$$ is definitely in Quadrant II.