Find the maximum value of the expression $\lfloor \tan^{-1} x - \tan^{-1} y \rfloor - \lfloor \sin^{-1} u - \sin^{-1} v \rfloor$,where $\lfloor . \rfloor$ denotes the greatest integer function,and $x, y, u, v$ are independent real variables.

If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$,then $\frac{d y}{d x}$ is equal to

If $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ and $\frac{-3\pi}{2} < x < \frac{-\pi}{2}$,then $\frac{dy}{dx} = $

For the least possible value of $n \in Z$,the solution $(x, y)$ of the equations $\cos ^{-1} x + (\sin ^{-1} y)^2 = \frac{n \pi^2}{4}$ and $(\cos ^{-1} x)(\sin ^{-1} y)^2 = \frac{\pi^4}{16}$ is

If $\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3 \cos^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the correct option$(s)$ is(are):
$(A) \cos \beta > 0$
$(B) \sin \beta < 0$
$(C) \cos(\alpha + \beta) > 0$
$(D) \cos \alpha < 0$

If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive integers,then the possible values of $(x, y)$ are: