Let $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, for $x \in \mathbb{R}$. Then the number of real solutions of the equation $\sqrt{1+\cos (2 x)}=\sqrt{2} \tan ^{-1}(\tan x)$ in the set $\left(-\frac{3 \pi}{2},-\frac{\pi}{2}\right) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ is equal to :

Considering only the principal values of the inverse trigonometric functions, the value of

$$ \frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^{2}}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^{2}}+\tan ^{-1} \frac{\sqrt{2}}{\pi} $$

is

The value of

$${\sec ^{ - 1}}\left( \matrix{ {1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) \hfill \cr} \right)$$

in the interval $$\left[ { - {\pi \over 4},\,{{3\pi } \over 4}} \right]$$ equals ..........

The number of real solutions of the equation $$\eqalign{ & {\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right) \cr & = {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right) \cr} $$ lying in the interval $$\left( { - {1 \over 2},{1 \over 2}} \right)$$ is ........... .

(Here, the inverse trigonometric functions sin^$$-$$1 x and cos^$$-$$1 x assume values in $${\left[ { - {\pi \over 2},{\pi \over 2}} \right]}$$ and $${\left[ {0,\pi } \right]}$$, respectively.)