The derivative of the function $f(x) = \cos^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x - 3\sin x) \right\} + \sin^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x + 3\sin x) \right\}$ with respect to $x$ at $x = \frac{3}{4}$ is:

If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$,then for $x < 0$ the value of $500 x^4+270 x^2+997=$

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

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

If $\alpha$ and $\beta$ (where $\alpha > \beta$) are two zeroes of the equation $3\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \pi$,then $(\alpha^2 + \beta^3)$ 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):
$(A) \cos \beta > 0$
$(B) \sin \beta < 0$
$(C) \cos(\alpha + \beta) > 0$
$(D) \cos \alpha < 0$