If $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $x \in (1, \infty)$,then $f'(x)$ is equal to:

The equation $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ has

If $2 \tan^{-1} x = 3 \sin^{-1} x$ and $x \neq 0$, then $8x^2 + 1 =$

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[-1, 1]$ such that $\cos ^{-1} x - \sin ^{-1} y = \alpha$,where $-\frac{\pi}{2} \leq \alpha \leq \pi$. Then,the minimum value of $x^2 + y^2 + 2xy \sin \alpha$ is

The value of $\cos^{-1}(\cos 12) - \sin^{-1}(\sin 14)$ 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):
$(A) \cos \beta > 0$
$(B) \sin \beta < 0$
$(C) \cos(\alpha + \beta) > 0$
$(D) \cos \alpha < 0$