 Hi, and welcome to the session. Let's work out the following question. The question says, prove that tan square a minus tan square b is equal to sin square a minus sin square b divided by cos square a into cos square b. Let's start with the solution. We see that the left-hand side is tan square a minus tan square b. This is equal to sin square a by cos square a minus sin square b upon cos square b. This is equal to sin square a into cos square b minus sin square b into cos square a and the whole divided by cos square a into cos square b. This is equal to sin square a into now cos square b can be written as 1 minus sin square b minus sin square b into cos square a can be written as 1 minus sin square a the whole divided by cos square a into cos square b. This is equal to sin square a minus sin square a into sin square b minus sin square b plus sin square b into sin square a and the whole divided by cos square a into cos square b. Now we see that minus sin square a into sin square b gets cancelled with plus sin square a into sin square b and we have sin square a minus sin square b divided by cos square a into cos square b which is also the RHS. So this is what we were supposed to prove in this question. I hope that you understood the solution and enjoyed the session. Have a good day.