Question:
Find the equation of the tangent to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\pi / 4$.
Solution:
finding slope of the tangent by differentiating $x$ and $y$ with respect to theta
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=1+\cos \theta$
$\frac{\mathrm{dy}}{\mathrm{d} \theta}=-\sin \theta$
Dividing both the above equations
$\frac{d y}{d x}=-\frac{\sin \theta}{1+\cos \theta}$
$m$ at theta $(\pi / 4)=-1+\frac{1}{\sqrt{2}}$
equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$
$y-1-\frac{1}{\sqrt{2}}=\left(-1+\frac{1}{\sqrt{2}}\right)\left(x-\frac{\pi}{4}-\frac{1}{\sqrt{2}}\right)$