On this problem, we want to remember that anytime we see a trigonometric inverse, that really really means it's an angle. So that sine inverse of two root 10 / 9 is really just Theta. So 2 sqrt 10 / 9 is sine of Theta. So if I draw a right triangle, my sine is my opposite side over hypotenuse. So 2 sqrt 10 / 9 I can find my third side. By the Pythagorean theorem we know the hypotenuse, so it's going to be the square root of the hypotenuse squared minus the known side squared. 9 ^2 is 81. Two sqrt 10 ^2 we have 2 * 2 is 4 and sqrt 10 times sqrt 10 is just 10. So 4 * 10 is 4081 -, 40. So now we want to find the secant of that Theta because remember we let that sine inverse of two root 10 / 9 all be Theta. So the secant of Theta secant is hypotenuse over adjacent. So we would get 9 / sqrt 41 and we want to rationalize that. So we'd get 9 square roots of 41 / 41.