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.
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sec (sin⁻¹ x/√x²+4)

Accepted Answer

Hello, everyone. We are asked to transform the following expression into an algebraic expression. We're going to use a right triangle in writing the algebraic expression. And we'll assume that the inverse trigonometric function is defined for its argument. And we'll also assume that X is greater than zero. We are given the C can of the arcs sign of four divided by the square root of in parentheses, X squared plus 16. We have four answer choices which are all slightly different algebraic expressions to begin with. It suggested that we work with the right triangle. So I'm gonna draw myself a right triangle. This way I can see a little clearer what I'm working with from there. I'm gonna take what's inside the parentheses first. So the arc sign of four divided by the square root of X squared plus 16. OK. I'm gonna make that equal to an angle. I'm gonna call Ceta and I'm gonna put the in my right triangle picture. I put it in the bottom right hand corner. So now I can take the sign of both sides and I'll have the sign of theta equals four divided by the square root of X squared plus 16. So I recall that when I was working with sine ratios, that sine is the opposite leg divided by the hypotenuse. So the like opposite theta will be a value of four. And our hypotenuse will have a value of the square root of X squared plus 16. We don't have our adjacent side. So I'm gonna label that a for now. And let's see when I do the can of theta, I will need to know the hypotenuse divided by the adjacent side. So let's find that adjacent side. Since this is a right triangle, we can use the Pythagorean theorem. So I'll use a squared plus the other leg which is four squared need ss equal the hypotenuse which is the square root of X squared plus 16 closed parentheses squared. We'll have a squared plus 16 equals and a square and a square root will cancel out. So X squared plus 16, subtracting 16 from both sides, I get the A squared equals X squared. And when I take the square root of both sides, this comes out to a equals X. So this means that our adjacent side has a value of X. So I'm gonna rewrite my cca so the can of theta will be our hypotenuse. So the square root of X squared plus divided by the adjacent leg which we said is X. So the C can of the, or the algebraic expression that goes with our trig expression is the square root of X squared plus 16, all of which is divided by X and that is answer choice C 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. ___ sec (sin⁻¹ x/√x²+4)

Key Concepts

Inverse Trigonometric Functions

Right Triangle Relationships

Secant Function

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