1. Evaluate tan[sin^-1(a)].

a. [sqrt(1-a^2)]/1-a^2
b. [sqrt(1-a^2)]/a
c. sqrt(1-a^2)
d. {a[sqrt(1-a^2)]}/1-a^2

I do not know the steps to find this answer and am not provided with a textbook. I have researched online some what, but I cannot seem to understand. Can someone please explain this?

draw the triangle. You want the angle whose sine is a = a/1

so, pick an acute angle, and its opposite side is a, and the hypotenuse is 1.
That means the adjacent side is √(1-a^2)

so, tan(arcsin(a)) = opposite/adjacent = a/√(1-a^2)

which, for those who insist on rational denominators, is (d)

right triangle sides a b c and angles A B C

C is right, c is hypotenuse
sin A = a/c
let c = 1
sin A = a
a^2 + b^2 = 1 so b = sqrt(1-a^2)
tan A = a/b = a/sqrt(1-a^2)
multiply top and bottom by sqrt(1-a^2) to get rid of sqrt in denominator
= a sqrt(1-a^2) / (1-a^2)

Oh my goodness. This makes so much more sense than the extremely confusing lesson this is in. Thank you Steve!

Well, evaluating trigonometric functions can sometimes be as confusing as trying to find your way out of a clown car! But don't worry, I'm here to bring some laughter and clarity to the situation.

To evaluate tan[sin^-1(a)], let's break it down step by step. We'll start with the inside function, sin^-1(a), which is also known as arcsine. So, arcsine returns the angle whose sine is a. Think of it as trying to find the angle a in a twisted game of hide-and-seek!

Now, if we look at the unit circle, we know that sin(theta) = opposite/hypotenuse. So, in our case, sin^-1(a) gives us the angle whose opposite side is a and hypotenuse is 1, which means the adjacent side is sqrt(1 - a^2)! It's like unraveling a mysterious clown trick!

Now that we have sin^-1(a) = theta, we can move on to evaluating tan[theta]. And tan(theta) is simply equal to opposite/adjacent. So, in our case, tan[sin^-1(a)] = a / sqrt(1 - a^2).

Now, let's look at the answer choices. None of them exactly match our result of a / sqrt(1 - a^2). But if we simplify it a bit, we can see that d) {a[sqrt(1-a^2)]}/1-a^2 is the closest match. It's like finding a clown nose that almost fits perfectly!

So, the correct answer is d) {a[sqrt(1-a^2)]}/1-a^2. Now you've successfully navigated the clown maze of trigonometry! Keep laughing and learning!

To evaluate tan[sin^(-1)(a)], we can make use of the trigonometric identity:

sin(sin^(-1)(x)) = x

This means that the sine of the inverse sine of a value is equivalent to the original value itself. In this case, we have:

sin(sin^(-1)(a)) = a

Now, to find tan[sin^(-1)(a)], we can use another trigonometric identity:

tan(x) = sin(x)/cos(x)

Since we have already determined that sin(sin^(-1)(a)) = a, we can substitute this into the equation:

tan[sin^(-1)(a)] = sin[sin^(-1)(a)] / cos[sin^(-1)(a)]

Substituting a into this equation, we get:

tan[sin^(-1)(a)] = a / cos[sin^(-1)(a)]

To evaluate the cosine of the inverse sine, we can use another trigonometric identity:

cos^2(x) + sin^2(x) = 1

Since sin(sine^(-1)(a)) = a, we can rewrite this equation as:

cos^2[sin^(-1)(a)] + a^2 = 1

Rearranging the equation, we can solve for cos[sin^(-1)(a)]:

cos[sin^(-1)(a)] = sqrt(1 - a^2)

Now substituting this expression back into the equation for tan[sin^(-1)(a)], we arrive at:

tan[sin^(-1)(a)] = a / sqrt(1 - a^2)

Therefore, the correct answer is option b) [sqrt(1-a^2)]/a.