Find the slope of the tangent or normal to given curve at the indicated points as instructed:

x= cos t, y = sin t at t = pi/4 : normal

x=y

$x= \cos t, y = \sin t$

$⇒\frac{dx}{dt}=-\sin t,⇒\frac{dy}{dt}=\cos t$

$⇒\frac{dy}{dx}=-\frac{\cos t}{\sin t}$

$⇒\frac{dy}{dx}=-\cot(t)$

$⇒\left.\frac{dy}{dx}\right]_{t=\frac{\pi}{4}}=-1$

∴ Slope of the tangent,

$⇒m=-\frac{1}{dy/dx}=+1$

∴ equation of the line → $x=y$