The possible values of $x$, which satisfy the trigonometric equation $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are

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

If $x = \sin (2 \tan^{-1} 2)$,$y = \cos (2 \tan^{-1} 3)$,and $z = \sec (2 \tan^{-1} 4)$,then:

Define $f: R \rightarrow R$ by $f(x) = \cos(\tan^{-1}(\sin(\tan^{-1} x)))$. Then $\lim_{x \rightarrow \infty} (f \circ f)(x)$ is equal to

Let $f:[0, 4\pi] \rightarrow [0, \pi]$ be defined by $f(x) = \cos^{-1}(\cos x)$. The number of points $x \in [0, 4\pi]$ satisfying the equation $f(x) = \frac{10-x}{10}$ is

If $\alpha$ and $\beta$ are roots of the equation $x^2+5|x|-6=0$,then the value of $|\tan^{-1} \alpha - \tan^{-1} \beta|$ is