The number of solutions of $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$ is

If the equation $\sin^{-1}(x - 1) + \cos^{-1}(x - 3) + \tan^{-1}\left(\frac{x}{-x^2 + 2}\right) = m$ holds,then the value of $m$ 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 the equation $\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x^2 - 1} + \tan^{-1} (\tan y) = a$ has at least one solution,then the number of integral values of $a$ is:

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:

$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} \, - \,\sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to