 I am welcome to the session. Let us discuss the following question. The question says, compute limit x tends to pi by 2 secant x minus tan x. Let's now begin with the solution. In this question, we have to evaluate limit of secant x minus tan x x tends to pi by 2. You know that secant x is equal to 1 by cos x and tan x is equal to sin x pi cos x. So this is equal to limit x tends to pi by 2 1 by cos x minus sin x by cos x. This is equal to limit y minus sin x divided by cos x as x tends to pi by 2. Now we will put h as x minus pi by 2. So as x tends to pi by 2, then h tends to 0. Now h equals to x minus pi by 2 implies x is equal to pi by 2 plus h. By substituting pi by 2 plus h in place of x, in this expression we get limit h tends to 0 1 minus sin pi by 2 plus h divided by cos pi by 2 plus h. Now this is equal to limit h tends to 0 1 minus cos h divided by minus sin h. This is equal to minus 1 into limit h tends to 0 1 minus cos h divided by sin h. This is equal to minus 1 into limit h tends to 0 1 minus cos h is 2 sin square h by 2 and sin h is 2 sin h by 2 cos h by 2. Now this is equal to minus 1 into limit h tends to 0 sin h by 2 divided by cos h by 2. This is equal to minus 1 into limit h tends to 0 tan h by 2. And this is equal to minus 1 into tan 0. Hence our required answer is 0 as tan 0 is equal to 1. So this completes the session. Bye and take care.