 Hello friends and how are you all doing today? The question says, find the equation of the tangent and normal to the given curves at the indicated points. We are given x equal to cos t, y equal to sin t, at t is equal to pi by 4. So, here first of all we will differentiate both x and y with respect to t and I am doing so we get dx by dt equal to d by dt of cos t. Here we have dy by dt equal to d by dt of sin t which gives us dx by dt t as derivative of cos t is equal to minus sin t that is dx by dt. And here we have dy by dt as the derivative of sin t which is cos t. Now, by chain rule that you must have started earlier in the chapters we have dy by dx equal to dy by dt divided by dx by dt. Which implies dy by dx is equal to dy by dt is cos t divided by minus sin t. Further implies dy by dx is equal to cos t upon sin t and having minus sin in the middle which will give us dy by dx equal to minus cot t. Now, at t is equal to pi by 4 we have the value of dy by dx at t is equal to pi by 4 minus cot pi by 4 whose value is minus 1. That means we have the slope of tangent that we take as m is minus 1 and slope of normal that is minus 1 upon m will be minus 1 upon minus 1 which will give us the value as 1. Now, with the help of the slopes we can find out equations. So, equation of tangent as well as normal y minus y1 equal to m x minus x1 where we have y minus y1 equal to minus 1 upon m x minus x1. Now, we need to have y1 and x1's value. So, first of all we know that we have x equal to cos t. So, x at the value of t that is pi by 4 we have cos pi by 4. That means the value of x is cos pi by 4 that is 1 by root 2. Similarly, we have y as sin pi by 4 that means y is equal to sin first of all we have y as sin t. On substituting the value of t as pi by 4 now we have the value of y as 1 upon root 2. So, now we have the slopes we have the value of x1 and y1. So, on substituting the values we have y minus 1 by root 2 equal to minus 1 x minus 1 by root 2. This is for tangent and y minus 1 by root 2 equal to 1 x minus 1 by root 2 is the equation of the norm. Let us simplify it we have y minus 1 by root 2 equal to minus x plus 1 by root 2. This implies x plus y minus 1 by root 2 minus 1 by root 2 gives us 2 root 2. So, 2 by root 2 which gives us x plus y minus root 2 is equal to 0. In the same manner we will be simplifying for equation of norm we have y minus 1 by root 2 equal to x minus 1 by root 2. Simplifying we have x is equal to y. So, the answer to the last and final fifth part is for tangent it is x plus y minus root 2 equal to 0 and for normal it is x is equal to y. This completes our session hope you understood the whole concept well do take care of your simplification and have a very nice day ahead.