 Hello and welcome to the session. In this session we are going to discuss the following question which says that evaluate tan of tan inverse of 4 by 5 plus cos inverse of 2 by 3 we know that tan inverse of x plus tan inverse of y is equal to tan inverse of x plus y upon 1 minus x y provided x y is less than 1 with this key idea let us proceed with the solution we are given the expression tan of tan inverse of 4 by 5 plus cos inverse of 2 by 3 here we shall convert sin inverse of 4 by 5 and cos inverse of 2 by 3 in terms of tan inverse function first of all let us assume sin inverse of 4 by 5 b theta which implies that 4 by 5 is equal to sin of theta or we can write sin theta is equal to 4 by 5 in a triangle ABC if theta is the angle between the lines AC and CB we know that sin theta is given by perpendicular upon hypotenuse that is AB by AC which will be equal to 4 by 5 now if perpendicular AB is given as 4 and hypotenuse AC is given as 5 then we can find the value of the base BC by using Pythagoras theorem we have the value of the base BC is given by square root of hypotenuse AC square minus perpendicular AB square which is equal to square root of 5 square minus 4 square that is square root of 25 minus 16 which is equal to square root of 9 that is 3 so we have base BC as 3 now we know that cos theta is given by base upon hypotenuse that is BC upon AC which is equal to 3 by 5 and tan of angle theta is given by sin theta by cos theta which is equal to 4 by 5 whole upon 3 by 5 therefore tan theta is given by 4 by 3 which implies that theta is equal to tan inverse of 4 by 3 now we have sin inverse of 4 by 5 equal to theta and theta is equal to tan inverse of 4 by 3 which implies that sin inverse of 4 by 5 is equal to tan inverse of 4 by 3 now again let us assume that cos inverse of 2 by 3 is equal to 5 which implies that 2 by 3 is equal to cos of 5 we can write cos of 5 is equal to 2 by 3 in a triangle PQR if 5 is the angle between the lines PR and RQ then cos of angle 5 is given by base upon hypotenuse that is RQ upon RP which is equal to 2 by 3 now if we are given the value of the base RQ as 2 and the value of the hypotenuse RP as 3 then we can find the value of the prependicular PQ by using Pythagoras theorem the Pythagoras theorem we have prependicular PQ is equal to square root of hypotenuse RP square minus base RQ square which is equal to square root of 3 square minus 2 square that is square root of 9 minus 4 which is equal to square root of 5 therefore the value of the prependicular PQ is square root of 5 and we know that sin of angle 5 is given by prependicular upon hypotenuse that is PQ upon PR which is equal to square root of 5 by 3 and tan of 5 is given by by cos of 5 which is equal to square root of 5 by 3 1 2 by 3 therefore tan of 5 is given by square root of 5 by 2 which implies that 5 is given by tan inverse of square root of 5 by 2 now we have cos inverse of 2 by 3 is equal to 5 and 5 is equal to tan inverse of square root of 5 by 2 which implies that cos inverse of 2 by 3 is equal to tan inverse of square root of 5 by 2 so now we have sin inverse of 4 by 5 is equal to tan inverse of 4 by 3 and cos inverse of 2 by 3 is equal to tan inverse of square root of 5 by 2 so we can replace sin inverse of 4 by 5 with tan inverse of 4 by 3 and cos inverse of 2 by 3 with tan inverse of square root of 5 by 2 in the given expression and hence we get tan of sin inverse of 4 by 5 plus cos inverse of 2 by 3 is equal to tan of tan inverse of 4 by 3 plus tan inverse of square root of 5 by 2 using the key idea we have tan inverse of x plus tan inverse of y is equal to tan inverse of x plus y upon 1 minus xy provided x into y is less than 1 here we assume 4 by 3 as x and square root of 5 by 2 as y so we can write tan of tan inverse of 4 by 3 plus square root of 5 by 2 upon 1 minus 4 by 3 into square root of 5 by 2 where x into y should be less than 1 that is 4 by 3 into square root of 5 by 2 should be less than 1 that is 2 into square root of 5 by 3 should be less than 1 which is true so we get tan of tan inverse of 8 plus 3 square root of 5 by 6 whole upon 1 minus 4 square root of 5 by 6 which is equal to tan of tan inverse of 8 plus 3 square root of 5 upon 6 minus 4 square root of 5 which is equal to 8 plus 3 square root of 5 upon 6 minus 4 square root of 5 on rationalizing this expression we get 8 plus 3 square root of 5 upon 6 minus 4 square root of 5 into 6 plus 4 square root of 5 upon 6 plus 4 square root of 5 which can be written as 8 plus 3 square root of 5 into 6 plus 4 square root of 5 whole upon 6 square minus 4 square root of 5 square which is equal to 48 plus 32 square root of 5 plus 18 square root of 5 plus 60 whole upon 26 minus 80 which is equal to 108 plus 50 square root of 5 upon minus of 34 which on further simplification can be written as minus of 54 plus 25 square root of 5 upon 22 therefore the value of the expression tan of sin inverse of 4 by 5 plus cos inverse of 2 by 3 is given by minus of 54 plus 25 square root of 5 whole upon 22 which is the required answer this completes our session hope you enjoyed this session see you next time