## Algebra and Trigonometry 10th Edition

True. It is because $csc~60°=\frac{1}{sin~60°}$
$sin~60°=\frac{\sqrt 3}{2}$ (See the table on page 436) $csc~60°=\frac{1}{sin~60°}=\frac{1}{\frac{\sqrt 3}{2}}=\frac{2}{\sqrt 3}$ So: $sin~60°csc~60°=\frac{\sqrt 3}{2}\frac{2}{\sqrt 3}=1$