If $k = \tan(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}(\frac{2}{3})) + \tan(\frac{1}{2}\sin^{-1}(\frac{2}{3}))$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is . . . . . . .

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:

The number of solutions of $\tan^{-1}\left(x+\frac{2}{x}\right) - \tan^{-1}\left(\frac{4}{x}\right) - \tan^{-1}\left(x-\frac{2}{x}\right) = 0$ is:

If $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$,then the derivative of $f(x)$ with respect to $g(x)$ is

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

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: