finding slope of the tangent by differentiating x and y with respect to theta
\(\frac{dx}{d\theta}=1+cos\theta\)
\(\frac{dy}{d\theta}=-sin\theta\)
Dividing both the above equations
\(\frac{dy}{dx}=-\frac{sin\theta}{1+cos\theta}\)
m(tangent) at theta ( \(\pi/2\) ) = – 1
normal is perpendicular to tangent so, m1m2 = – 1
m(normal) at theta ( \(\pi/2\) ) = 1
equation of tangent is given by y – y1 = m(tangent)(x – x1)
\(y-1=-1(x-\frac{\pi}{2}-1)\)
equation of normal is given by y – y1 = m(normal)(x – x1)
\(y-1=1(x-\frac{\pi}{2}-1)\)