What is the angle of inclination of the tangent to the curve $y = (x - 1)(x - 2)$ at the point $(1, 0)$ with the $x$-axis?

Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$,where $x \in (0, \frac{\pi}{2})$. $A$ normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point:

At what point on the curve ${x^3} - 8{a^2}y = 0$ is the slope of the normal equal to $\frac{-2}{3}$?

The equation of the normal to the curve $2x^{2} + 3y^{2} - 5 = 0$ at the point $P(1, 1)$ is:

If the tangent at a point $P$ on the curve $y=4x^4+x$ is perpendicular to the tangent to the same curve at $(0,0)$,then the point $P$ is

If $\beta$ is an angle between the normals drawn to the curve $x^2+3y^2=9$ at the points $P(3 \cos \theta, \sqrt{3} \sin \theta)$ and $Q(-3 \sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \frac{\pi}{2})$,then: