 Hello and welcome to the session. Let's work out the following question. It says prove the following identity where the angles involved are acute angles for which the expression is defined. So let's now move on to the solution. The given expression is on the left hand side sine theta minus 2 sine cube theta upon 2 cos cube theta minus cos theta. We have to prove that this expression is equal to tan theta. We take sine theta common from the numerator so we have 1 minus 2 sine square theta upon taking cos theta common from the denominator. We have cos theta into 2 cos square theta minus 1. Now sine theta upon cos theta is tan theta minus 2 sine square theta. Sine square theta can be written as 1 minus cos square theta as we know that sine square theta plus cos square theta is 1. So sine square theta is equal to 1 minus cos square theta upon 2 cos square theta minus 1. So again this is equal to tan theta into 1 minus 2 minus 2 into minus 2 into minus cos square theta is plus 2 cos square theta upon 2 cos square theta minus 1. Now again this is equal to tan theta into 1 minus 2 is minus 1 plus 2 cos square theta that is 2 cos square theta minus 1 upon 2 cos square theta minus 1. Now 2 cos square theta minus 1 gets cancelled with 2 cos square theta minus 1 and we are left with tan theta and that is the RHS. That is what we have to prove. Hence proved. So this completes the question and the session. Bye for now. Take care. Have a good day.