 Hello and welcome to the session. Let's work out the following question. Let's say it's proved 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 and we have to prove this identity using the identity cosikin squared a is equal to 1 plus cot squared a. Now let's start with LHS which is equal to cos a minus sin a plus 1 upon cos a plus sin a minus 1. Now since we have to use the identity cosikin squared a is equal to 1 plus cot squared a, we need to convert the whole expression in terms of cosikin a and cot a. So we divide the numerator and denominator by sin a. So we have cos a upon sin a minus sin a upon sin a plus 1 upon sin a upon cos a upon sin a plus 1 upon sin a minus 1 upon sin a. Now cos a upon sin a is cot a sin a divided by sin a is 1 plus 1 upon sin a is cosikin a upon cot a plus cosikin a minus plus 1 since here we have sin a. So sin a divided by sin a is 1 and here we have minus cosikin a 1 upon sin a is cosikin a again this is equal to cot a plus cosikin a minus 1 upon cot a minus cosikin a plus 1. Now we will multiply the numerator and denominator by cot a minus cosikin a. Now multiplying cot a minus cosikin a with cot a plus cosikin a we will be having cot squared a minus cosikin squared a multiplying cot a minus cosikin a with minus 1 we have minus cot a minus cosikin a upon cot a minus cosikin a plus 1 into cot a minus cosikin a. Here we have used the formula of a plus b into a minus b is a square minus b square in this case a is cot a and b is cosikin a. Now cot square a minus cosikin square a minus 1 minus of cot a minus cosikin a upon cot a minus cosikin a plus 1 into cot a plus cot a minus cosikin a here we have used the formula cot square a minus cosikin square a is equal to minus 1 now again this is equal to minus 1 minus cot a plus cosikin a upon cot a minus cosikin a plus 1 into cot a minus cosikin a now taking minus common from the numerator we have 1 plus cot a minus cosikin a upon 1 plus cot a minus cosikin a into cot a minus cosikin a now this gets cancelled with this and we are left with minus 1 upon cot a minus cosikin a now again cot a can be written as cos way upon sin a cosikin a can be written as 1 upon sin a taking LCM we have cos a minus 1 upon sin a and this is further equal to minus sin a upon cos a minus 1 taking minus common from the denominator we have sin a upon 1 minus cos a now multiplying numerator and denominator by 1 plus cos a we have sin a into 1 plus cos a upon 1 minus cos a into 1 plus cos a is 1 minus cos square a further equal to sin a into 1 plus cos a upon 1 minus cos square a is sin square a so this is equal to 1 plus cos a upon sin a now we will solve the RHS RHS is cosikin a plus cot a now cosikin a can be written as 1 plus cot square a under the root of 1 plus cot square a plus cot a as we know that cosikin square a is equal to 1 plus cot square a now again cot square a can be written as cos square a upon sin square a plus cot a that is cos a upon sin a taking LCM we have square root of sin square a plus cos square a upon sin square a plus cos a upon sin a now we know that sin square a plus cos square a is 1 so we have square root of 1 upon sin square a plus cos a upon sin a now this is equal to 1 upon sin a now LCM is sin a and in the numerator we have 1 plus cos a which is same as LHS we have proved above that LHS is 1 plus cos a upon sin a and we have simplified RHS and we have obtained RHS is equal to LHS hence the identity is proved so this completes the question on the session why for now take care have a good day