At any point ‘$t$’ on the curve $x=a(t+\sin t),$ $y=a(1-\cos t)$, find the lengths of tangent, normal, subtangent an subnormal. 

Given curves $x=a(t+\sin t);y=a(1-\cos t)$

diff w.r. to ‘$t$’

$\frac{dx}{dt}=a[1+\cos t]\frac{dy}{dt}=a[0-(-\sin t\left)\right]$

$=a2{\cos }^{2}t/2=a\sin t$

$=2a{\cos }^{2}t/2=a(2\sin t/2 \cos t/2)$

$m=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a\sin t/2\cos t/2}{2a{\cos }^{2}t/2}=\tan t/2$

Length of tangent $=\left|\frac{y}{m}\sqrt{1+m^2}\right|$

$=\left|\frac{a(1-\cos t)}{\tan t/2}\sqrt{1+{\tan }^{2}t/2}\right|$

$=\left|a\frac{2{\sin }^{2}t/2}{\frac{\sin t/2}{\cos t/2}}\sqrt{{\sec }^{2}t/2}\right|$

$=|2a\sin t/2\cos t/2(\sec t/2\left)\right|$

$=\left|2a\sin t/2\cos t/2\frac{1}{\cos t/2}\right|=|2a\sin t/2|=2a\sin t/2$

Length of normal $=\left|y\sqrt{1+m^2}\right|$

$=\left|a(1-\cos t)\sqrt{1+{\tan }^{2}t/2}\right|=\left|a2{\sin }^{2}t/2\sqrt{{\sec }^{2}t/2}\right|$

$=|2a\sin t/2 \sin t/2 \sec t/2|$

$=\left|2a\sin t/2\frac{\sin t/2}{\cos t/2}\right|=|2a\sin t/2\tan t/2|$

Length of sub tangent $=\left|\frac{y}{m}\right|$

$=\left|\frac{a(1-\cos t)}{\tan t/2}\right|=\left|\frac{a2{\sin }^{2}t/2}{\frac{\sin t/2}{\cos t/2}}\right|=|2a\sin t/2\cos t/2|$

$=\left|a\cdot \sin 2\left(\frac{t}{2}\right)\right|=|a\cdot \sin t|$

Length of sub normal $=\left|y\mathrm{m}\right|$

$\left.=\left|a\right(1-\cos t)\tan t/2|=\mid a2{\sin }^{2}t/2\right)\tan t/2\mid$

$=\left|2a{\sin }^{2}t/2\tan t/2\right|$