## Calculus (3rd Edition)

$$\sin \theta =c, \quad \tan\theta =\frac{c}{\sqrt{1-c^2}}, \quad \csc \theta =\frac{1}{c}.$$
By making use of the Pythagorean theorem, we have $$\sin \theta =\frac{c}{1}=c, \quad \tan\theta =\frac{c}{\sqrt{1-c^2}}$$ $$\csc \theta =\frac{1}{\sin\theta}=\frac{1}{c}.$$