To express [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex], follow these steps:
1. Understand the function [tex]\(\cos^{-1} x\)[/tex]:
- [tex]\(\cos^{-1} x\)[/tex] (or arccosine of [tex]\(x\)[/tex]) is the angle [tex]\(\theta\)[/tex] such that [tex]\(\cos(\theta) = x\)[/tex].
- Thus, denote [tex]\(\theta = \cos^{-1} x\)[/tex]. Consequently, [tex]\(\cos(\theta) = x\)[/tex].
2. Express [tex]\(\cot(\theta)\)[/tex] in terms of trigonometric functions:
- The cotangent function is given by: [tex]\(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\)[/tex].
3. Substitute [tex]\(\cos(\theta)\)[/tex]:
- Since [tex]\(\cos(\theta) = x\)[/tex], we need to find [tex]\(\sin(\theta)\)[/tex].
4. Use the Pythagorean identity for sine:
- The identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex] helps here.
- Substitute [tex]\(\cos(\theta) = x\)[/tex] into the identity: [tex]\(\sin^2(\theta) + x^2 = 1\)[/tex].
- Solving for [tex]\(\sin^2(\theta)\)[/tex]: [tex]\(\sin^2(\theta) = 1 - x^2\)[/tex].
- Then, [tex]\(\sin(\theta) = \sqrt{1 - x^2}\)[/tex] (we take the positive root because [tex]\(\theta = \cos^{-1} x\)[/tex] gives an angle in the range [tex]\([0, \pi]\)[/tex], where sine is non-negative).
5. Combine the expressions:
- Now, substitute [tex]\(\cos(\theta) = x\)[/tex] and [tex]\(\sin(\theta) = \sqrt{1 - x^2}\)[/tex] into the cotangent expression:
[tex]\[
\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{x}{\sqrt{1 - x^2}}
\][/tex]
Hence, the expression [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex] is:
[tex]\[
\cot \left(\cos^{-1} x\right) = \frac{x}{\sqrt{1 - x^2}}
\][/tex]
This completes the step-by-step process of rewriting [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex].
