Hey everyone. In this problem, we're asked to evaluate the expression the inverse tangent of negative square root of 3. Now, when working with inverse trigonometric functions, we know that we can also think of this as, okay, the tangent of what angle is equal to negative square root 3, and we want to find the angle for which that is true. Now looking at our unit circle here, we know that when working with the inverse tangent, our value has to be between negative pi over 2 and pi over 2, which is actually the exact same interval that we dealt with when working with the inverse sine. Now again here, we could get even more specific because all of my tangent values in quadrant 1 are going to be positive whereas all of my tangent values in quadrant 4 are going to be negative. So if I'm taking the inverse tangent of a positive value, I know my solution has to be an angle in quadrant 1. Whereas if I'm taking the inverse tangent of a negative value, my answer has to be an angle in quadrant 4. Now here, since I'm taking the inverse tangent of the negative square root of 3, I am looking for an angle in this bottom quadrant, quadrant 4. So for which of these angles is the tangent negative square root of 3? Well, for negative pi over 3, since this is a reference angle to pi over 3, I know that the tangent of this value is going to be negative square root of 3. So that represents my solution, negative pi over 3. So this is my solution. The inverse tangent of negative root 3 is tan-1−√3 equals −π3, and we are good to go here. Thanks so much for watching, and I'll see you in the next one.