Explanation

(c)
Given curve is
$x = a(t + \sin t),\ y = a(1 - \cos{t)}$
$\Rightarrow \ \ \frac{dx}{dt} = a\left( 1 + \cos t \right),\frac{dy}{dt} = a\left( \sin t \right)$
$\therefore\ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\left( \sin t \right)}{a\left( 1 + \cos t \right)}$
$= \frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\cos^{2}\frac{t}{2}} = \tan\frac{t}{2}$
Length of the normal$= y\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}$
$= a\left( 1 - \cos t \right)\sqrt{1 + \tan^{2}\left( \frac{t}{2} \right)}$
$= a\left( 1 - \cos t \right)\sec\left( \frac{t}{2} \right)$
$= 2a\sin^{2}\left( \frac{t}{2} \right)\sec\left( \frac{t}{2} \right)$
$= 2a\sin\left( \frac{t}{2} \right)\tan\left( \frac{t}{2} \right)$