The inclination of the tangent at $\theta =\frac{\pi }{3}$ on the curve $x=a(\theta +\sin \theta ), y=a(1+\cos \theta ) is$

The given curve is $x=a(\theta +\sin \theta ), y=a(1+\cos \theta )$

The slope of the tangent is $\frac{dy}{dx}$ at $\theta =\frac{\mathrm{\pi }}{3}$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}\\ & =& \frac{a\left(-\sin \theta \right)}{a\left(1+\cos \theta \right)}\\ & =& -\frac{\sin \theta }{1+\cos \theta }\end{array}$

Hence, 

  $\begin{array}{rcl}m& =& -\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{3}{2}}}\\ & =& -\frac{1}{\sqrt{3}}\end{array}$

Therefore, the inclinaiton is 

 $\begin{array}{rcl}{\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)& =& \mathrm{\pi }-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ & =& \mathrm{\pi }-\frac{\mathrm{\pi }}{6}\\ & =& \boxed{\frac{5\mathrm{\pi }}{6}}\end{array}$