Everyone, in this problem, we're asked to evaluate the expression, the tangent of the inverse sine of \( \frac{x}{\sqrt{x^2 + 4}} \). Now, this problem might look really overwhelming because there are multiple variables, a square root, and a bunch of stuff going on. But we can still attack this the same way we would any of these composite trig functions that are definitely not on our unit circle. And \( \frac{x}{\sqrt{x^2 + 4}} \), definitely not on there. So let's go ahead and get started with step number 1.
Now for step 1, we're going to take that inside function, and we're going to use our interval to determine our quadrant. Now here, looking at this inside function, I am faced with finding the inverse sine. And now when working with the inverse sine, I know that I am working with angles from negative \( \frac{\pi}{2} \) to positive \( \frac{\pi}{2} \). So my triangle either has to be in quadrant 4 or quadrant 1. Now looking at the sine of our argument here, our argument is kind of strange, but we can still take a look at its sign. \( \frac{x}{\sqrt{x^2 + 4}} \) is a positive value, so that means that my triangle has to go in quadrant 1 or my sine values are positive. So step 1 is done. We can move on to step number 2, which is to draw our triangle and use our argument in order to label our angle theta and our two sides. So up here in quadrant 1, let's go ahead and draw our right triangle.
And with my right triangle, I'm going to label my angle theta right here. And with this argument, this tells me that the sine of my angle theta is equal to \( \frac{x}{\sqrt{x^2 + 4}} \). This seems strange, but hang with me here. Using SOHCAHTOA, this tells me that my opposite side has to be \( x \), and my hypotenuse is \( \sqrt{x^2 + 4} \). So we're still working through this the same way. We've completed step number 2. We can move on to step number 3, where we're going to use the Pythagorean theorem to find that missing third side. So let's go ahead and set our Pythagorean theorem up here. We have \( a^2 + b^2 = c^2 \), and we're going to be solving for one of our leg lengths, so either \( a \) or \( b \).
Here, I'm going to solve for \( a \). So I'm going to keep \( a \) as variable, \( a^2 \), and then plus my \( b^2 \). \( B \) here is \( x \). I'm squaring that value. And that's equal to \( c^2 \). Now \( c \) here is this whole expression, \( \sqrt{x^2 + 4} \), and all of that gets squared. Now from here, let's simplify this algebraically. Now here on my left side, I can't really do anything. I'm just left with \( a^2 + x^2 \). But on that right side, if I square this, I'm going to get rid of that square root. So I'm just left with \( x^2 + 4 \). Now from here, if I'm solving for \( a \), I want to go ahead and isolate \( a \) by subtracting \( x^2 \) from both sides.
Now when I do that, it cancels on that left side, but it also cancels on that right side. So I'm simply left with \( a^2 = 4 \). This looks like a much more friendly expression. So here, we're just left to take the square root of both sides. We're left with \( a = \sqrt{4} \), which is 2. So our missing side length here is 2. Now we've completed step number 3. We can move on to our final step, which is to use our triangle in order to evaluate that outside function. Now, here, our outside function is the tangent. And in order to find the tangent of our angle theta, using SOHCAHTOA, we're going to take our opposite side. We're going to divide it by our adjacent side.
We have all of that information. Here, our opposite side is \( x \). \( X \). Our adjacent side is 2. So my final answer here is \( \frac{x}{2} \). The tangent of the inverse sine of \( \frac{x}{\sqrt{x^2 + 4}} \) is equal to \( \frac{x}{2} \). And we're completely done here. Thanks for watching, and I'll see you in the next one.