$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) = $

$\tan \left(\cos ^{-1} \frac{1}{\sqrt{2}}+\tan ^{-1} \frac{1}{2}\right) = $

If for some $\alpha, \beta$ such that $\alpha \leq \beta$ and $\alpha+\beta=8$,the equation $\sec^2(\tan^{-1} \alpha) + \operatorname{cosec}^2(\cot^{-1} \beta) = 36$ holds,then the value of $\alpha^2+\beta$ is . . . . . .

The solution set of the inequality $(\tan^{-1} x)(\cot^{-1} x) - (\tan^{-1} x)(1 + \frac{\pi}{2}) - 2\cot^{-1} x + 2(1 + \frac{\pi}{2}) > \lim_{x \to \infty} [\sec^{-1} x - \frac{\pi}{2}]$ is (where $[.]$ denotes the greatest integer function):

Let $x = \frac{m}{n}$ ($m, n$ are co-prime natural numbers) be a solution of the equation $\cos(2 \sin^{-1} x) = \frac{1}{9}$ and let $\alpha, \beta$ $(\alpha > \beta)$ be the roots of the equation $mx^2 - nx - m + n = 0$. Then the point $(\alpha, \beta)$ lies on the line

Let $x = \sin(2 \tan^{-1} \alpha)$ and $y = \sin(\frac{1}{2} \tan^{-1} \frac{4}{3})$. If $S = \{\alpha \in R : y^2 = 1 - x\}$,then $\sum_{\alpha \in S} 16 \alpha^3$ is equal to $...........$