Question

Let O be an angle in quadrant II such that cos0= -8/9Find the exact values of csc0 and cot0Csc0=Cot0=

Answer 1

Given the angle, θ is an angle in quadrant II

`\cos \theta=-(8)/(9)`

From the given equation:

`\cos \theta=(adjacent)/(hypotenuse)`

So, the adjacent = 8 and the hypotenuse = 9

Using the Pythagorean to find the opposite side

`\text{opposite}=\sqrt[]{9^2-8^2}=\sqrt[]{17}`

we will find csc θ and cot θ

so,

`\begin{gathered} \csc \theta=(hypotenuse)/(opposite)=\frac{9}{\sqrt[]{17}}=\frac{9\sqrt[]{17}}{17} \\ \\ \cot \theta=(adjacent)/(opposite)=-\frac{8}{\sqrt[]{17}}=-\frac{8\sqrt[]{17}}{17} \end{gathered}`