Given $\theta=\pi / 3$
and $x=a(\theta-\sin \theta), y=a(1-\cos \theta)$
Differentiate it, w.r.t. ' $\theta$
$$
\frac{d x}{d \theta}=a(1-\cos \theta), \frac{d y}{d \theta}=a(\sin \theta)
$$

Then, $\frac{d y}{d x}=\frac{a \sin \theta}{a(1-\cos \theta)}=\frac{\sin \theta}{1-\cos \theta}$
$$
\left(\frac{d y}{d x}\right)_{\theta=\pi / 3}=\frac{\sin \pi / 3}{1-\cos \pi / 3}=\frac{\sqrt{3} / 2}{1-1 / 2}=\sqrt{3}
$$
$\therefore$ Slope of tangent $=\sqrt{3}$
$$
\begin{aligned}
& \text { Length of subtangent }=\frac{(y)_{\theta=\pi / 3}}{m} \\
&=\frac{a(1-\cos \pi / 3)}{\sqrt{3}}=\frac{a(1 / 2)}{\sqrt{3}}=\frac{a}{2 \sqrt{3}}
\end{aligned}
$$

Length of subnormal
$$
\begin{aligned}
& =(y)_{\theta=\pi / 3} \cdot m=a(1-\cos \pi / 3) \cdot \sqrt{3} \\
& =a\left(1-\frac{1}{2}\right) \sqrt{3}=\frac{a \sqrt{3}}{2}
\end{aligned}
$$

Sum of Eqs. (i) and (ii),
$$
\begin{aligned}
\frac{a}{2 \sqrt{3}}+\frac{a \sqrt{3}}{2} & =\frac{\sqrt{3} a}{6}+\frac{a \sqrt{3}}{2}=\frac{\sqrt{3} a+3 \sqrt{3} a}{6} \\
& =\frac{4 \sqrt{3} a}{6}=\frac{2 \sqrt{3} a}{3}=\frac{2 a}{\sqrt{3}}
\end{aligned}
$$
$\therefore$ Sum of length of subtangent and sub-normal is $\frac{2 a}{\sqrt{3}}$.