Question

Use a right triangle to write the following expression in algebraic expression. Assume that x is positive and in the domain of the given inverse trigonometric function.tan(cos−1(9x))=

Answer 1

Answer:

The algebraic expression for $tan(cos^{-1}(9x))=\frac{\sqrt{1-81x^{2}}}{9x}$

Step-by-step explanation:

Let θ = $cos^{-1}(9x)$ use the properties of inverse trigonometric functions $cos(\theta)=cos(cos^{-1}(\theta))=\theta\\cos(\theta)=cos(cos^{-1}(9x))=9x$

In a right angled triangle, the cosine of an angle is

$cos(\theta)=\frac{adjacent}{hypotenuse}$

Use this expression $cos(\theta)=9x$  to find what are the sides of the right triangle.

$cos(\theta)=\frac{adjacent}{hypotenuse}\\cos(\theta)=\frac{9x}{1}$

Next find what is the expression for the opposite side, for this use the Pythagorean theorem and the values above

$opposite^{2}+adjacent^{2}=hypotenuse^{2}\\opposite^{2}= hypotenuse^{2}-adjacent^{2}\\opposite =\sqrt{1-81x^{2}}$

We said that $\theta = cos^{-1}(9x)$, so now we can use the definition of tangent $tangent(\theta)=\frac{opposite}{adjacent}$ and the right triangle that we defined to find the algebraic expression for

$tan(cos^{-1}(9x))$

$tan(cos^{-1}(9x))=tan(\theta) \\ tan(\theta)=\frac{\sqrt{1-81x^{2}}}{9x}$