Find the slopes of the tangent and the normal to the curve `x=a(theta-sintheta),\ \ y=a(1-costheta)` at `theta=pi//2`

Answer

`m=(-frac{d x}{d y})=-1`

Explanation

The slope of normal `x=a(\theta-\sin \theta) ; \quad y=a(1-\cos \theta)` at `\theta=\frac{\pi}{2}`
$$ \left(\frac{\mathrm{dx}}{\mathrm{d} \theta}\right)_{\theta=\frac{\pi}{2}}=\mathrm{a}(1-\cos \theta)=\mathrm{a} $$
$$
\begin{aligned}
&\left(\frac{d y}{d \theta}\right)_{\theta=\frac{\pi}{2}}= a\sin \theta =a \\
&\therefore\left(\frac{d y}{d x}\right)=1
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