 Hello friends and how are you all doing today? We need to evaluate the following. This is the fourth part. Let's quickly proceed with the solution. Here we will be writing down the values of sin 30 tan 45 cos 30 cos 60 cos 45 from the table that is known to us. Now let's proceed. We have sin 30 as 1 by root sorry 1 by 2 plus tan 45 as 1 minus cos x 60 as 2 by root 3. It is getting divided by sin 30 which is 2 root 3 plus cos 60 which is 1 by 2 plus cot 45 degree that is 1. Let's simplify taking 2 root 3 as the L sin we are left with root 3 plus 2 root 3 minus 4 in the numerator. Here also 2 root 3 will be our L sin we have 4 plus root 3 plus 2 root 3. Simply find we have root 3 plus 2 root 3 as 3 root 3 minus 4 it is getting divided by 3 root 3 plus 4. We will be rationalizing the denominator by the conjugate of 3 root 3 plus 4 that is 3 root 3 minus 4. Now here if you carefully observe we have 3 root 3 minus 4 into 3 root 3 minus 4 which will be 3 root 3 minus 4 the whole square and in the denominator we have 8 plus b getting multiplied by a minus b where a is 3 root 3 and b is 4. So here we will have a square that is 3 root 3 square minus b square which is 4 square this is because a plus b into a minus b gives us a square minus b square and we know that a minus b the whole square is equal to a square plus b square minus 2 a b right. So we will be using this identity over here to solve it we have a square plus b square minus 2 a into b upon where our a is 3 root 3 and b is 4 upon simplifying it we have 27 minus 16. Now we have 27 plus 16 minus on simplifying we have 2 multiplied by 3 gives us 6 6 multiplied by 4 gives us 24 and we have root 3 along with it upon 27 minus 16 gives us 11. So we have the answer the required answer now as 27 plus 16 gives us 43 we have minus 24 by root 3 with it upon 11 and this is the required answer to the fourth and final part. If you understood the solution well do remember the identity that you learnt in your previous classes and have a very nice day ahead.