Answer
The equation of the tangent line at $t=\frac{\pi }{4}$ is
$y=\left(1+\sqrt{2}\right) x+2-\frac{\pi}{4} \left(1+\sqrt{2}\right)$.
Work Step by Step
The cycloid generated by the unit circle has parametric equations:
$x(t)=t-\sin t$, $y(t)=1-\cos t$.
Using Eq. (8) the slope of the tangent line is
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{\sin {t}}{1-\cos {t}}$.
At $t=\frac{\pi }{4}$, the slope of the tangent line is $\frac{dy}{dx}|_{t=\frac{\pi }{4}}=1+\sqrt{2}$.
The corresponding point is
$c(\frac{\pi}{4})=\left(-\frac{\sqrt{2}}{2}+\frac{\pi }{4},1-\frac{\sqrt{2}}{2}\right)$
So, the equation of the tangent line at $t=\frac{\pi }{4}$ is
$y-\left(1-\frac{\sqrt{2}}{2}\right)=\left(1+\sqrt{2}\right) \left(x-\left(-\frac{\sqrt{2}}{2}+\frac{\pi }{4}\right)\right)$.
Or,
$y=\left(1+\sqrt{2}\right) x+2-\frac{\pi}{4} \left(1+\sqrt{2}\right)$.
