Answer:
The algebraic expression for [tex]tan(cos^{-1}(9x))=\frac{\sqrt{1-81x^{2}}}{9x}[/tex]
Step-by-step explanation:
Let θ = [tex]cos^{-1}(9x)[/tex] use the properties of inverse trigonometric functions [tex]cos(\theta)=cos(cos^{-1}(\theta))=\theta\\cos(\theta)=cos(cos^{-1}(9x))=9x[/tex]
In a right angled triangle, the cosine of an angle is
[tex]cos(\theta)=\frac{adjacent}{hypotenuse}[/tex]
Use this expression [tex]cos(\theta)=9x[/tex] to find what are the sides of the right triangle.
[tex]cos(\theta)=\frac{adjacent}{hypotenuse}\\cos(\theta)=\frac{9x}{1}[/tex]
Next find what is the expression for the opposite side, for this use the Pythagorean theorem and the values above
[tex]opposite^{2}+adjacent^{2}=hypotenuse^{2}\\opposite^{2}= hypotenuse^{2}-adjacent^{2}\\opposite =\sqrt{1-81x^{2}}[/tex]
We said that [tex]\theta = cos^{-1}(9x)[/tex], so now we can use the definition of tangent [tex]tangent(\theta)=\frac{opposite}{adjacent}[/tex] and the right triangle that we defined to find the algebraic expression for
[tex]tan(cos^{-1}(9x))[/tex]
[tex]tan(cos^{-1}(9x))=tan(\theta) \\ tan(\theta)=\frac{\sqrt{1-81x^{2}}}{9x}[/tex]
