$\cot \left\{\frac{2019 \pi}{2}-\left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)\right\}$ is equal to:

The complete solution set of the inequality $(\sec^{-1}x - 4)(\sec^{-1}x - 1)(\sec^{-1}x - 2) \ge 0$ is

Which pair$(s)$ of function$(s)$ is/are equal? (where ${x}$ and $[x]$ denote the fractional part and integral part functions respectively.)

Let $g(x) = f(x) + f(1-x)$ and $f''(x) > 0$ for $x \in (0, 1)$. If $g$ is decreasing in the interval $(0, \alpha)$ and increasing in the interval $(\alpha, 1)$,then $\tan^{-1}(2\alpha) + \tan^{-1}\left(\frac{1}{\alpha}\right) + \tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)$ is equal to:

If $x \in [0, 1]$,then the number of solution$(s)$ of the equation $2[\cos^{-1}x] + 6[\text{sgn}(\sin x)] = 3$ is (where $[.]$ denotes the greatest integer function and $\text{sgn}(x)$ denotes the signum function of $x$)-

The value of $x$, where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is