The angle made by the tangent of the curve $x = a$ ($t + \sin t$ cost); $y = a(1 + \sin t)^2$ with the $x$-axis at any point on it is

For the parametric curve, differentiate with respect to $t$. $$x=a(t+\sin t\cos t)$$ $$y=a(1+\sin t)^2$$ So, $$\frac{dx}{dt}=a(1+\cos 2t)$$ $$\frac{dy}{dt}=2a(1+\sin t)\cos t$$ Hence the slope of the tangent is $$\frac{dy}{dx}=\frac{2a(1+\sin t)\cos t}{a(1+\cos 2t)}$$ Using $1+\cos 2t=2\cos^2 t$, $$\frac{dy}{dx}=\frac{1+\sin t}{\cos t}$$ Now, $$\frac{1+\sin t}{\cos t}=\frac{\left(\sin \frac{t}{2}+\cos \frac{t}{2}\right)^2}{\left(\cos \frac{t}{2}+\sin \frac{t}{2}\right)\left(\cos \frac{t}{2}-\sin \frac{t}{2}\right)}$$ $$\frac{1+\sin t}{\cos t}=\frac{\cos \frac{t}{2}+\sin \frac{t}{2}}{\cos \frac{t}{2}-\sin \frac{t}{2}}$$ $$\frac{dy}{dx}=\tan \left(\frac{\pi}{4}+\frac{t}{2}\right)$$ Therefore the angle $\theta$ made by the tangent with the $x$-axis is $$\theta=\frac{\pi}{4}+\frac{t}{2}=\frac{1}{4}(\pi+2t)$$ On checking the options, this matches option $A$.