 Hello and welcome to the session. In this session we are going to discuss the following question which says that evaluate sin of cos inverse of square root of 5 by 3 plus sin inverse of 5 by 13. We know that sin of theta plus phi is equal to sin of theta cos of phi plus cos of theta sin of phi. With this key idea let us proceed with the solution. We have the expression sin of cos inverse of square root of 5 by 3 plus sin inverse of 5 by 13. Let theta be equal to cos inverse of square root of 5 by 3 which implies that cos theta is equal to square root of 5 by 3 In a triangle ABC if theta is the angle between the lines AC and CB. And we know that cos of triangle theta is given by base upon hypotenuse that is BC upon AC which will be equal to square root of 5 by 3. Now if we have base BC equal to square root of root 5 and hypotenuse AC equal to 3 then we can find the value of the prependicular AB by using Pythagoras theorem. By Pythagoras theorem we have prependicular AB is equal to square root of hypotenuse AC square minus base BC square which is equal to square root of 3 square minus square root of 5 square which will be equal to square root of 9 minus 5 which is given by square root of 4 that is 2 value of the prependicular AB as we know that sin of theta is equal to prependicular upon hypotenuse that is AB by AC which is equal to 2 by 3 therefore the theta as square root of 5 by 3 theta equal to 2 by 3. Now again let phi be equal to sin inverse of 5 by 13 which implies that is equal to 5 by 13 triangle PQR if phi is the angle between the lines PR and RQ. We know that sin of angle 5 is given by prependicular upon hypotenuse that is PQ by PR which is equal to 5 by 13 at the value of the prependicular PQ as 5 and the value of the hypotenuse PR as 13 then we can find the value of the base RQ by using Pythagoras theorem. By Pythagoras theorem we have RQ is given by square root of hypotenuse PR square minus prependicular PQ square which is equal to square root of 13 square minus 5 square which will be square root of 169 minus 25 that is square root of 144 given by 12 therefore base RQ is equal to 12. Now we know that cos of angle 5 is given by base upon hypotenuse that is RQ upon RP which is equal to 12 by 13 therefore we have sin of angle 5 as 5 by 13 and cos of angle 5 as 12 by 14. We have assumed that cos inverse of square root of 5 by 3 as theta and sin inverse of 5 by 13 as 5 so the given expression becomes sin of theta plus 5 which is given by sin theta cos 5 plus cos theta sin 5 that is equal to sin theta cos 5 plus cos theta sin 5 now substituting the value of cos theta sin theta and cos 5 sin 5 on the right hand side of the equation we get 2 by 3 into 12 by 13 plus square root of 5 by 3 into 5 by 13 which is equal to 24 by 39 plus 5 square root of 5 by 39 that is 24 plus 5 square root of 5 whole upon 39 so sin of is equal to 24 plus 5 square root of 5 whole upon 39. Therefore we can write the value of the expression sin of cos inverse of square root of 5 by 3 plus sin inverse of 5 by 13 is given by 24 plus 5 square root of 5 whole upon 39 which is the required answer. This completes our session. Hope you enjoyed this session.