Question

Please help me the answer has to be written in simplified rationalized form

Answer 1

We are given the following trigonometric ratio

`\csc \theta=\frac{\sqrt[]{11}}{3}`

We are asked to find cosθ

Recall that cscθ and sinθ are related as

`\sin \theta=(1)/(\csc\theta)=\frac{1}{\frac{\sqrt[]{11}}{3}}=\frac{3}{\sqrt[]{11}}`

Also, recall that sinθ is equal to

`\sin \theta=(opposite)/(hypotenuse)=\frac{3}{\sqrt[]{11}}`

This means that

Opposite = 3

Hypotenuse = √11

Let us find the adjacent side using the Pythagorean theorem

`\begin{gathered} (adjacent)^2+(opposite)^2=(hypotenuse)^2 \\ (adjacent)^2+(3)^2=(\sqrt[]{11})^2 \\ (adjacent)^2=(\sqrt[]{11})^2-\mleft(3\mright)^2 \\ (adjacent)^2=11-9 \\ (adjacent)^2=2 \\ √((adjacent)^2)=√(2) \\ adjacent=\sqrt[]{2} \end{gathered}`

So, the adjacent side is √2

Finally, cosθ is given by

`\cos \theta=(adjacent)/(hypotenuse)=\frac{\sqrt[]{2}}{\sqrt[]{11}}*\frac{\sqrt[]{11}}{\sqrt[]{11}}=\frac{\sqrt[]{2}\cdot\sqrt[]{11}}{11}=\frac{\sqrt[]{22}}{11}`

Therefore, cosθ = √22/11

`\cos \theta=\frac{\sqrt[]{22}}{11}`