Find the equation of normal to the curve `x=cos theta, y = sin theta" at point "theta = pi/4*`

Explanation

At `theta = pi/4`
`x=cos pi/4 = 1/sqrt2 and y = sin. pi/4 = 1/sqrt2`
`:." Co-ordinates of the point "=(1/sqrt2,1/sqrt2)`
Now `(dx)/(d theta)=-sin theta and (dy)/(d theta) = cos theta`
`rArr (dy)/(dx) = (dy//d theta)/(dx//d theta) = (cos theta)/(-sin theta) =- cot theta`
At `theta = pi//4," Slope of tangent " m_(1) =- cot pi/4 =-1`
and slope of normal `m_(2) = (-1)/(m_(1))=1`
`:." Equation of normal at " theta = pi/4 ," i.e., at the point "(1/sqrt2,1/sqrt2)*`
`y-1/sqrt2 = 1 (x-1/sqrt2)`
`rArr x-y = 0`.