If $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,then $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ is equal to $........$.

Consider the following statements:
Assertion $(A)$: When $x, y, z$ are positive numbers, then $\operatorname{Tan}^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right) = \pi$
Reason $(R)$: $\operatorname{Tan}^{-1} a + \operatorname{Tan}^{-1} b = \operatorname{Tan}^{-1}\left(\frac{a+b}{1-ab}\right)$ if $a > 0$ and $b > 0$ and $ab < 1$.

If $\theta = \tan^{-1} a$,$\phi = \tan^{-1} b$ and $ab = -1$,then $\theta - \phi = $

If $y = \tan^{-1}\sqrt{\frac{1 + \cos x}{1 - \cos x}}$,then $\frac{dy}{dx}$ is equal to

Evaluate: $\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = $

Find $\frac{dy}{dx}$ if $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$,where $0 < x < \frac{1}{\sqrt{2}}$.