## Trigonometry 7th Edition

RECALL: $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} \\\csc{\theta} = \dfrac{1}{\sin{\theta}}$ Thus, the left side of the equation is equivalent to: $\require{cancel} \dfrac{\sin{\theta}}{\tan{\theta}}$ $\require{cancel}\\=\dfrac{\sin{\theta}}{\frac{\sin{\theta}}{\cos{\theta}}} \\=\sin{\theta} \cdot \dfrac{\cos{\theta}}{\sin{\theta}} \\=\cancel{\sin{\theta}} \cdot \dfrac{\cos{\theta}}{\cancel{\sin{\theta}}} \\=\cos{\theta}$ With $\dfrac{\sin{\theta}}{\tan{\theta}}=\cos{\theta}$, the left side of the given equation is not equal to the right side. The statement is false.