 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says, prove that the function f given by fx is equal to log cos x is strictly decreasing on an open interval 0 to pi by 2 and strictly increasing on an open interval pi by 2 to pi. So let's start the solution. Given fx is equal to log cos x, now f dash x is equal to 1 over cos x into minus sin x and this is equal to minus tan x. Now for 0 less than x less than pi by 2, now tan x is positive in the first quadrant therefore minus tan x is negative. Therefore f dash x is less than 0 is strictly decreasing on an open interval 0 to pi by 2. Now for less than pi this is negative in the second quadrant hence minus tan x because tan x is negative in second quadrant. Therefore our f dash x is greater than 0. This implies f is increasing on an open interval pi by 2 to pi. Hence a given function fx is equal to log cos x is strictly decreasing on an open interval 0 to pi by 2 and strictly increasing on an open interval pi by 2 to pi. Hence proved. I hope the solution is clear to you. Bye and take care.