 Hello and welcome to the session. In this session we are going to discuss the following question which says that proves that cos inverse of 7 by 18 plus sin inverse of 6 by 10 is equal to cos inverse of 28 minus 3 into square root of 315 by 90. We know that cos inverse of x plus cos inverse of y is equal to cos inverse of xy minus square root of 1 minus x square into square root of 1 minus y square. If both x and y are greater than 0 and x square plus y square is less than equal to 1, with this key idea we shall move on to the solution. We need to prove that cos inverse of 7 by 18 plus sin inverse of 6 by 10 is equal to cos inverse of 28 minus 3 into square root of 315 by 90. Taking the left hand side of the expression we have cos inverse of 7 by 18 plus sin inverse of 6 by 10. Now we shall convert sin inverse of 6 by 10 in terms of cos inverse function. Let theta be equal to sin inverse of 6 by 10 which implies that sin of theta is equal to 6 by 10 in a triangle ABC if theta is the angle between the lines AC and CB. We know that sin of theta is given by perpendicular upon hypotenuse that is AB upon AC which will be equal to 6 by 10. That is if we have the value of the perpendicular AB as 6 and the value of the hypotenuse AC as 10 then we can find the value of the base CB by using the Pythagoras theorem. By Pythagoras theorem we have base BC is equal to square root of hypotenuse BC square minus perpendicular AB square that is equal to square root of 10 square minus 6 square which is given by square root of 100 minus 36. That is equal to square root of 64 which is given by 8. So base BC is equal to 8. Now we know that cos of theta is equal to base upon hypotenuse that is BC upon AC which is equal to 8 by 10. Which implies that theta is equal to cos inverse of 8 by 10. So here we have sin inverse of 6 by 10 as theta and theta is equal to cos inverse of 8 by 10. Which implies that sin inverse of 6 by 10 is equal to cos inverse of 8 by 10. So we can replace sin inverse of 6 by 10 with cos inverse of 8 by 10 in the left hand side of the given expression. Therefore the left hand side becomes cos inverse of 7 by 18 plus cos inverse of 8 by 10. Now from the key idea we have cos inverse of x plus cos inverse of y is equal to cos inverse of xy minus square root of 1 minus x square into square root of 1 minus y square provided both x and y are greater than 0 and x square plus y square is less than equal to 1. Here we assume x is 7 by 18 and y is 8 by 10 and we have 7 by 18 is greater than 0 and 8 by 10 is greater than 0 also 7 by 18 square plus 8 by 10 square is less than 1. Therefore we can write the value of the expression cos inverse of 7 by 18 plus cos inverse of 8 by 10 is equal to cos inverse of 7 by 18 into 8 by 10 minus square root of 1 minus 7 by 18 square into square root of 1 minus 8 by 10 square. Which is equal to cos inverse of 14 upon 35 minus square root of 1 minus 39 upon 364 into square root of 1 minus 64 by 100 which further gives cos inverse of 14 by 35 minus square root of 364. Minus 39 upon 364 into square root of 100 minus 64 by 100 which is equal to cos inverse of 14 by 35 minus square root of 315 by 364 into square root of 36 by 100 which is equal to cos inverse of 8 by 10 square. Cos inverse of 14 by 35 minus square root of 315 by 18 into 6 by 10 which is further solving this cos inverse of 13 by 35 minus square root of 315 by 30. Now on taking the LPN written in the brackets we get cos inverse of 28 minus 3 into square root of 315 by 90. So we have cos inverse of 7 by 18 plus cos inverse of 8 by 10 is equal to cos inverse of 28 minus 3 into square root of 315 by 90. Or we can also write it as cos inverse of 7 by 18 plus sin inverse of 6 by 10 is equal to cos inverse of 28 minus 3 into square root of 315 by 90 which is equal to the right hand side. Hence we have plus the result that is cos inverse of 7 by 18 plus sin inverse of 6 by 10 is equal to cos inverse of 28 minus 3 square root of 315 by 90. This completes our session but you enjoy this session.