The coordintes of the point P on the curve $\mathrm{X}=\mathrm{a}(\mathrm{\theta }+\mathrm{sin\theta }), \mathrm{y}=(1-\mathrm{cos\theta })$ where the tangent is inclined at an angle $\frac{\mathrm{\pi }}{4}$ to the x-axis, are

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \tan \frac{\mathrm{\pi }}{4}=1\\ \mathrm{x}& =& \mathrm{a}\left(\mathrm{\theta }+\sin \mathrm{\theta }\right) \mathrm{y}=\mathrm{a}(1-\cos \mathrm{\theta })\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{\mathrm{a}(0+\sin \mathrm{\theta })}{\mathrm{a}(1+\cos \mathrm{\theta })}=\frac{2\sin {\displaystyle \frac{\mathrm{\theta }}{2}} \cos \frac{\mathrm{\theta }}{2}}{2{\cos }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}=\tan \frac{\mathrm{\theta }}{2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& 1 \Rightarrow \tan \frac{\mathrm{\theta }}{2}=1 \Rightarrow \frac{\mathrm{\theta }}{2}=\frac{\mathrm{\pi }}{4} \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{2}\\ \mathrm{x}& =& \mathrm{a}\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1\right) \mathrm{y}=\mathrm{a}(1-0)=\mathrm{a}\end{array}$

$\therefore$ Point $(\mathrm{x}, \mathrm{y})=\left(\mathrm{a}\left(\frac{\mathrm{\pi }}{2}+1\right), \mathrm{a}\right)$