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 at theta ( \(\pi/4\) ) = \(-1+\frac{1}{\sqrt{2}}\)
equation of tangent is given by y – y1 = m(tangent)(x – x1)
\(y-1-\frac{1}{\sqrt{2}}\)\(=(-1+\frac{1}{\sqrt{2}})(x-\frac{\pi}{\sqrt{4}}-\frac{1}{\sqrt{2}})\)
