 Hello and welcome to the session. In this session we are going to discuss the following question which says that proof that sec of tan inverse of whose sec of cot inverse of 1 by x is equal to square root of 2x square plus 1 by x. Let us proceed with the solution. We are given the expression sec of tan inverse of to sec of cot inverse of 1 by x here. Let theta be equal to cot inverse of 1 by x which implies that cot of theta is equal to 1 upon x. In a triangle ABC, if theta is the angle between the lines AC and CB, then cot of angle theta is given by base upon BC upon OB which is equal to 1 upon x. Now, if base BC is given by 1 and perpendicular OB is given by x, then we can find the hypotenuse AC by using the Pythagoras theorem. So, by Pythagoras theorem we have hypotenuse AC is equal to square root of square of the perpendicular OB plus square of the base BC which is equal to square root of x squared plus 1 squared that is 1. Now we know that cot of angle theta is given by hypotenuse by perpendicular that is AC upon OB which is given by square root of x squared plus 1 by x. Therefore we have cot of angle theta is equal to square root of x squared plus 1 by x which implies that theta is given by cot inverse of square root of x squared plus 1 by x. We have assumed the value of theta as cot inverse of 1 by x and theta is equal to cot inverse of square root of x squared plus 1 by x. Therefore we can say that cot inverse of 1 by x is equal to cot inverse of square root of x squared plus 1 by x. Therefore the expression cot inverse of 1 by x becomes cot inverse of square root of x squared plus 1 by x which is equal to square root of x squared plus 1 by x. Now putting the value of cot inverse of 1 by x as square root of x squared plus 1 upon x in the given expression we get cot inverse of cot inverse of 1 by x is square root of x squared plus 1 by x. Now let phi be equal to tan inverse of square root of x squared plus 1 by x which implies that tan of phi is equal to square root of x squared plus 1 by x. Now in a triangle PQR if phi is the angle between the lines PR and RQ then tan of angle phi is given by perpendicular upon base that is PQ upon RQ which is given by square root of x squared plus 1 by x. If perpendicular PQ is given by square root of x squared plus 1 and base RQ is given by x then we can find the value of the hypotenuse PR by using the Pythagoras theorem. By Pythagoras theorem we have hypotenuse PR is equal to square root of square of the perpendicular PQ plus square of the base RQ which is equal to square root of PQ squared that is square root of x squared plus 1 by whole squared which is equal to x squared plus 1 plus RQ squared that is x squared which is equal to square root of 2x squared plus 1. So hypotenuse PR is equal to square root of 2x squared plus 1. Now we know that sec of angle phi is given by hypotenuse by base that is RP upon RQ which is equal to square root of 2x squared plus 1 by x. Therefore sec of angle phi is equal to square root of 2x squared plus 1 by x which implies that phi is equal to 2x squared plus 1 upon x. Now as we have assumed the value of phi as tan inverse of square root of x squared plus 1 by x and phi is also equal to sec inverse of square root of 2x squared plus 1 by x. This implies that tan inverse of square root of x squared plus 1 by x is equal to sec inverse of square root of 2x squared plus 1 by x. Therefore the expression sec of tan inverse of square root of x squared plus 1 by x can be written as sec of second inverse of square root of 2x squared plus 1 by x which is equal to square root of 2x squared plus 1 by x hence proved as a result that is sec of tan inverse of good sec of cot inverse of 1 by x is equal to square root of 2x squared plus 1 by x. This completes our session. Hope you enjoyed this session.