 Hello and welcome to the session. Let's work out the following question. It says prove that y equal to 4 sin theta upon 2 plus cos theta minus theta is an increasing function of theta in the closed interval 0 to pi by 2. So let's now move on to the solution. Before moving on to the solution we should first know the criteria for an increasing function. A function is increasing in an open interval if its derivative is greater than 0 in that interval. So we'll show that the derivative of this function is greater than 0 in the open interval 0 to pi by 2. Now the given function is 4 sin theta upon 2 plus cos theta minus theta. Now we'll differentiate this with respect to theta. So dy by d theta equal to we'll differentiate this with quotient rule. So we have 2 plus cos theta as it is into derivative of 4 sin theta which is 4 cos theta minus 4 sin theta as it is into derivative of 2 plus cos theta which is minus sin theta upon 2 plus cos theta whole square minus derivative of theta with respect to theta is 1. Now again this is equal to 4 into 2 cos theta plus cos square theta plus sin square theta we have taken 4 common upon 2 plus cos theta whole square minus 1. Now again this is equal to 4 into 2 cos theta plus 1 upon 2 plus cos theta whole square minus 1. Again taking asian simplifying we have 4 into 2 cos theta plus 1 minus 2 plus cos theta whole square will be 4 plus cos square theta plus 4 cos theta upon 2 plus cos theta whole square. Simplifying we have 4 into 2 cos theta is 8 cos theta plus 4 minus 4 minus cos square theta minus 4 cos theta upon 2 plus cos theta whole square. This is equal to 4 cos theta minus cos square theta upon 2 plus cos theta whole square. So this is equal to cos theta into 4 minus cos theta upon 2 plus cos theta whole square. Now we have to show that dy by d theta is greater than 0 in the open interval 0 to pi by 2. So for theta lying between 0 pi by 2 dy by d theta which is equal to cos theta into 4 minus cos theta upon 2 plus cos theta whole square. Now in the open interval 0 to pi by 2 cos theta is greater than 0. 4 minus cos theta is also greater than 0 because theta lies between 0 to pi by 2 therefore cos theta lies between 0 and 1 right. So 4 minus cos theta will be strictly greater than 0 and cos the denominator is greater than 0 being a perfect square dy by d theta which is equal to cos theta into 4 minus cos theta upon 2 plus cos theta whole square is greater than 0 for theta lying between 0 to pi by 2. Hence y is increasing function of theta close interval 0 to pi by 2. Hence the result is this completes the question and the session. Hope you enjoyed the session. Bye for now. Take care. Have a good day.