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.

csc (cot⁻¹ x)

Accepted Answer

Hello, everyone. We are asked to transform the following expression into an algebraic expression. We're gonna use a right triangle in writing the algebraic expression. And we will assume that the inverse trigonometric function is defined for its argument. And we will also assume that X is greater than zero. The expression we are given is the can of the arc tangent of three X. There are four answer choices, all of which are slightly different algebraic expressions to begin with it stated we should use a right triangle. So I'm gonna draw myself a right triangle as reference. I will fill in what I find as I go. So the co C can of the arc tangent of three X. I'm gonna start with what's in the parentheses. So the arc tangent of three X, I'm gonna set that equal to the angle theta theta represents any angle. We don't know which one. And in my image, I'm gonna put it in the bottom right hand corner. So that will be my angle theta. Now I can take the tangent of both sides. So I get the tangent of theta equals three X. I recall that my tangent ratio is the opposite side divided by the adjacent side or the opposite leg divided by the adjacent leg. So three X would be the same as saying three X divided by one. So now I know the leg opposite theta on my right triangle has a value of three X and the leg adjacent to theta has a value of one. Ultimately, I need to find the of theta and this will be the hypotenuse divided by the opposite leg. Well, at the moment, we don't know the value for the hypotenuse. So I'm gonna call that A and I'm gonna use the Pythagorean theorem to find the value of A. So Pythagorean theorem says my legs squared. So three X in parentheses squared plus one squared will equal the hypotenuse squared. So A squared, three X squared is nine X squared plus one, squared is one and that will equal A squared taking the square to both sides. I find that A equals the square root of nine X squared plus one. So all of that is under a radical. So it's the square root of nine X squared plus one. So now that I know what A is, I know the hypotenuse, I can go back the can of alpha. I'm sorry of theta will be the hypotenuse, which in this case is the square root of nine X squared plus one divided by this like opposite theta, which is three X. So that is actually our new expression. It is the square root of nine X squared plus one, all of which is divided by three X and that matches answer choice A have a nice day.

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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. csc (cot⁻¹ x)