Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. The secant function F of X equals secant X. It is a function because it passes the vertical line test. The domain is all reals where X is not equal to π halves plus some K Pi. The range negative Infinity -1 union one to Infinity. When we look at the secant function. If we want to find the inverse secant, we can realize that the secant function is not currently 1 to one. We need to figure out how to restrict it so that we have all possible Y values that occur. We want it to be as close together as possible, and we always want to include the angle 0. So if we look at it, we want this angle 0, if we think off to the right are going to be all the positive YS. To get all the negative YS, we're going to need to do from Pi halves to π so F inverse of X equals secant inverse X. The domain is going to be the range of the original. So negative Infinity to -1 union one to Infinity, the range is going to be our restriction. Our restriction being zero to π halves and Pi halves to π. We can't include Pi halves because that's an asymptote. If we look at our cosecant, our cosecant function F of X equal cosecant X is going to have a domain of all reals where X doesn't equal K π. The range again is negative Infinity to -1 union one to Infinity. So the inverse cosecant, currently the cosecant is not one to one. We need to restrict it and we want to use positive Y. We need to use all the positive YS and all the negative YS. So here's the positive Y, here's the negative Y, and we're going to use negative π halves to 0, union zero to π halves. We don't include that zero because there is an asymptote that's going to be the range for the inverse, where the domain is the negative Infinity to -1 union one to Infinity. If we look at the cotangent function, we have the domain being all real numbers, where X isn't equal to K Pi. We're going to use our range negative Infinity to Infinity the inverse cotangent. We want to cover all the values and we want to do it in this case continuously SO1 full. So in between 0 and π is going to be our restriction for inverse. So zero to π is going to be the range. The domain is the range of the original function, hence negative Infinity to Infinity. So looking at cotangent inverse of negative root 3, we want to remember that we need an angle that's in between 0 and π here. So when is cotangent inverse negative root 3? That's going to be true at 5 Pi 6. For cosecant inverse, it's going to be in between negative π halves to Pi halves where it can never equal 0. So -2 root 3 / 3 in between negative π halves to Pi halves without it equaling 0 is going to occur at the angle negative π thirds. This is from knowing our unit circle secant inverse of -2. Remember, secant has to have our angle measure between zero and π where Theta is not going to equal π halves. So seek an inverse of -2 in between 0 and π would give us out 2π thirds. Now we're going to be able to combine a bunch of things. If I wanted the cosine inverse of negative root 3 / 2, cosine inverse has got to be in between 0 and π. So what angle gives us out negative root 3 / 2 for cosine that's in between 0 and π and that answer is going to be 5 Pi 6. Now we want to figure out what is the cosecant of five Pi six. Well, cosecant is the inverse of sine, so sine of five Pi 6 is positive 1/2 or cosecant of five Pi 6 is 2. Once again, if we know our unit circle, we can get cosec 5 Pi six straight off of there. This next one, cotangent of 2π thirds, cotangent 2π thirds, that's going to be a negative root 3 / 3. And now we're asking when is tangent inverse negative root 3 / 3. Remember, tangent has got to be in between negative π halves and Pi halves. So when we're looking at this one, we know we want to have negative root 3 / 3 for tangent is going to occur at negative Pi 6. Looking at some more examples, what if it wasn't exactly on our unit circle? Well, we know tangent inverse of -3 has got to be in between negative π halves and Pi halves. If we thought about putting a triangle in our unit circle grid idea, we know that the -3 wood forest us being into the 4th quadrant and we also know that tangents the opposite side over the adjacent side. So now we know two sides of a right triangle. Can we find the third side? Yes, we're going to use Pythagorean theorem. So we have 1 ^2 + 3 ^2 or sqrt 10. So now we have this triangle and we want the sine value of this triangle. So the sine of whatever this angle is, let's call it Theta, is going to be a negative because it's in the 4th quadrant for sine, and sine is opposite over the hypotenuse. Now, we never leave a radical in the denominator, so we're going to rationalize it. So our answer is going to be -3 root 10 / 10. Well, what if we had variables instead? Cotangent is going to be in between 0 and π. So in this case, we are going to have first possibly or second, but the sign in both the 1st and 2nd is going to be positive. So it doesn't matter which one of those quadrants I use, we're just going to go with quadrant one. If it's cotangent, cotangent is adjacent over opposite. So one goes on the opposite and U goes on the adjacent because U is the same thing as U / 1. Can we find that third side? Yes, going to use Pythagorean Theorem 1 ^2 + U ^2 or sqrt 1 + U ^2. Now the sign in both the 1st and the second quadrants is positive, so this is going to equal the opposite side over the hypotenuse. We always have to rationalize, so we're going to multiply by sqrt 1 + U ^2 over sqrt 1 + U ^2 and we're going to get 1 + U ^2 / 1 + U ^2. What if we had cosine of secant inverse U? Well, secant is restricted. Remember between 0 and π where theta's not equaling Pi halves. So here, if we're thinking about the 1st and the second quadrant, well, secant's positive in the 1st, but it's negative in the second. However, cosine is also positive in the 1st and negative in the second. So those signs are going to match whatever we do. Secant is going to be hypotenuse over adjacent. Can we find our third side? Yes. This time we knew our hypotenuse, so we're going to subtract. So we're going to get U ^2 -, 1. When I look at the cosine now, the cosine is going to be the adjacent over the hypotenuse. Oh, we didn't even need to find that third side. So 1 / U. Looking at some more, if we're given some functions F of X equals sine XG of X equal cosine XH of X equal tangent X with the appropriate restrictions, can we find G inverse of F of seven Pi force? And the answer is going to be yes, we're going to put in seven Pi force into the F of X equation. So what is sine of seven Pi force? Sine of seven Pi force is going to be negative root 2 / 2. Now we want to figure out the inverse of the G function. So we want cosine inverse of negative root 2 / 2. Well what angle within our cosine restriction? Remember our cosine restriction is zero to π gives us out negative root 2 / 2 and our answer is going to be 3 Pi force. What if we had one such as the next example where we have F inverse of 2/5 so we want H of sine inverse of -2 fifths. The sine is restricted between negative π halves to Pi halves. If we want it to be negative, we know it's got to be in the 4th quadrant, sine being opposite over hypotenuse. If I know two sides, can we find the third one? We know the hypotenuse and we know one of the sides. So 5 ^2 -, 2 ^2 or sqrt 21. Now we want the H of that. So let's call this angle in here Theta. So we want tangent of that Theta. Well, if we're looking at tangent of the Theta, tangent in the 4th quadrant is negative and tangent is opposite over adjacent. We're going to rationalize that and we're going to get -2 sqrt 21 / 21. Thank you and have a wonderful day.