Answer:
[tex]\displaystyle \csc(\cos^{-1}(x))=\frac{1}{\sqrt{1-x^2}}[/tex]
Step-by-step explanation:
Please refer to the attachment.
We have:
[tex]\csc(\cos^{-1}(x))[/tex]
First, we will let:
[tex]\cos^{-1}(x)=\theta[/tex]
Then:
[tex]x=\cos(\theta)[/tex]
So, the adjacent side of our triangle is x and the hypotenuse is 1.
Then by the Pythagorean Theorem, the opposite side is given by:
[tex]x^2+o^2=1^2[/tex]
So:
[tex]o=\sqrt{1-x^2}[/tex]
Going back, we have:
[tex]\csc(\cos^{-1}(x))[/tex]
Since arccos(x) is θ:
[tex]=\csc(\theta)[/tex]
Cosecant is the ratio of the hypotenuse over the opposite side. Therefore:
[tex]\displaystyle \csc(\theta)=\frac{1}{\sqrt{1-x^2}}[/tex]
Hence:
[tex]\displaystyle \csc(\cos^{-1}(x))=\frac{1}{\sqrt{1-x^2}}[/tex]
