If $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$,where $-1 \le x \le 1$,$-2 \le y \le 2$,and $x \le \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

Let $\alpha = 3 \sin^{-1}(\frac{6}{11})$ and $\beta = 3 \cos^{-1}(\frac{4}{9})$, where inverse trigonometric functions take only the principal values. Given below are two statements:
Statement $I$: $\cos(\alpha + \beta) > 0$.
Statement $II$: $\cos(\alpha) < 0$.
In the light of the above statements, choose the correct answer from the options given below:

Show that $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$

If $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$,then $\frac{dy}{dx}$ is

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

The sum of the values of $x$ satisfying the equation $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1}(x)$ is: