If $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$, then $x^2=$

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

The total number of real solutions of the equation $\theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)$ is $($Here,the inverse trigonometric functions $\sin ^{-1} x$ and $\tan ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,respectively.$)$

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:

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

The value of $\sec ^2(\tan ^{-1} 2)+\operatorname{cosec}^2(\cot ^{-1} 3)$ is