If cot𝜽= 𝟒𝟎/𝟗, find the values of cosecθ and sinθ.

Solution We know that $$\cot(\theta) = \text{adjacent}/\text{opposite}$$ in a right triangle, where $$\theta$$ is one of the acute angles of the triangle. Given $$\cot(\theta) = 40/9$$, we can label the adjacent side of the right triangle as 40 and the opposite side as 9. Using the Pythagorean theorem, we can find the hypotenuse: $$a^2 + b^2 = c^2$$ $$9^2 + 40^2 = c^2$$ $$81 + 1600 = c^2$$ $$1681 = c^2$$ $$c = \sqrt{1681} = 41$$ Now we can use the definitions of the trigonometric functions to find the values of $$\csc(\theta)$$ and $$\sin(\theta)$$: $$\csc(\theta) = \text{hypotenuse}/\text{opposite} = 41/9$$ $$\sin(\theta) = \text{opposite}/\text{hypotenuse} = 9/41$$ Therefore, $$\csc(\theta) = 41/9$$ and $$\sin(\theta) = 9/41$$.