In Exercises 83–94, use a right triangle to write each expression...

Question

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x.tan (cos⁻¹ x)

Accepted Answer

Hello, everyone. We are asked to transform the following expression into an algebraic expression. We can use a right triangle in writing the algebraic expression. And assume that the inverse trigonometric function is defined for its argument. We will also assume that X is greater than zero. The expression we are given is the tangent of the arc sign of two X. We have four answer choices which are varying algebraic expressions to begin with. I'm gonna draw a right triangle. So I have something to reference and from there, I'm gonna take what's in the parentheses. So the arc sign of two XX and set it equal to an angle theta. I'm gonna put data on my right triangle. I put it in the bottom right hand corner, it doesn't matter which angle you call theta as long as you're consistent with wi referencing it. So now the arc sign of two X equals data, I can take the sign of both sides and I'll get the sign of theta equals two X. Since we're working with the right triangle, I recall that in right triangle trig, the ratio for sine is the opposite leg divided by the hypotenuse here we have two X, which we can treat as two X divided by one. So this means the leg opposite the theta angle is two X. And our hypotenuse has a value of one. Once we find the value of the, we would need to do the tangent of theta. So if we're gonna do the tangent of theta thinking to those ratios, this would be the opposite leg divided by the adjacent like, well, I don't know the adjacent leg yet. So we need to solve it. I'm gonna call our missing leg. A and then we'll use the Pythagorean theorem. So the sum of the squares of legs. So a squared plus in parentheses, two X squared has to equal the hypotenuse squared. So one squared. So A squared plus four X squared will equal one subtracting four X squared from both sides. As we are trying to solve for A, we have a squared equals one minus four X squared. I'm gonna take the square root of both sides and make sure I take the square root of the whole side. And that will give me that a equals the square root of one minus four X squared. So I'm putting that on my right triangle. So now that I know both the opposite and adjacent sides, let's see if I can find the tangent of data. So the tangent of data again is the opposite side. So two X divided by the adjacent, which is the square root of one minus four X squared. Well, we need to rationalize that denominator. We shouldn't leave it with a radical in the denominator. So we're gonna multiply both the numerator and denominator by the square root of one minus four X squared. And when we do so, we will get two X multiplied by the square root of one minus four X squared divided by one minus four X squared. That looks like as simplified as this can be. So it is our answer and it matches answer choice B have a nice day.

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x.tan (cos⁻¹ x)

Key Concepts

Inverse Trigonometric Functions

Right Triangle Relationships

Trigonometric Identities

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