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

$\sin \theta=\frac{1}{\csc\theta}=\frac{1}{\frac{\sqrt[]{11}}{3}}=\frac{3}{\sqrt[]{11}}$

Also, recall that sinθ is equal to

$\sin \theta=\frac{opposite}{hypotenuse}=\frac{3}{\sqrt[]{11}}$

This means that

Opposite = 3

Hypotenuse = √11

Let us find the adjacent side using the Pythagorean theorem

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

So, the adjacent side is √2

Finally, cosθ is given by

$\cos \theta=\frac{adjacent}{hypotenuse}=\frac{\sqrt[]{2}}{\sqrt[]{11}}\times\frac{\sqrt[]{11}}{\sqrt[]{11}}=\frac{\sqrt[]{2}\cdot\sqrt[]{11}}{11}=\frac{\sqrt[]{22}}{11}$

Therefore, cosθ = √22/11

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