$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} \, - \,\sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to

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

If $x, y, z$ are in arithmetic progression and $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are also in arithmetic progression,then:

If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$,then $\frac{d y}{d x}$ is equal to

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):

If ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi ,$ then the value of $a\sqrt {1 - {a^2}} + b\sqrt {1 - {b^2}} + c\sqrt {1 - {c^2}}$ is