 Hello and welcome to the session. Let us understand the following problem today. Prove that tan inverse of under root 1 plus x minus under root 1 minus x whole divided by under root 1 plus x plus under root 1 minus x is equal to pi by 4 minus half cos inverse x where minus 1 by root 2 is less than equal to x is less than equal to 1. Here we will be using is equal to cos 2 theta and identity as 1 plus cos 2 theta is equal to 2 cos square theta and 1 minus cos 2 theta is equal to 2 sin square theta 4 minus theta is equal to 1 minus tan theta by 1 plus tan theta. This is our key idea. Now let us write the solution. Put x is equal to cos it implies theta is equal to half cos inverse of x inverse of under root 1 plus x minus under root 1 minus x whole divided by under root 1 plus x plus under root 1 minus x which is equal to tan inverse of under root 1 plus cos 2 theta minus under root 1 minus cos 2 theta whole divided by under root 1 plus cos 2 theta plus under root 1 minus cos 2 theta which is equal to tan inverse of under root 2 cos square theta minus under root 2 sin square theta whole divided by under root 2 cos square theta plus under root 2 sin square theta. This is we get from the key idea which is equal to tan inverse of cos theta minus sin theta by cos theta plus sin theta. Now dividing numerator and denominator by cos theta we get tan inverse of 1 minus tan theta by 1 plus tan theta which is equal to tan inverse of tan of pi by 4 minus theta using identity. This is equal to pi by 4 minus theta which is equal to pi by 4 minus half cos inverse which is equal to RHS. Therefore LHS is equal to RHS hence proved. I hope you understood the problem. Bye and have a nice day.