If $\cot \left(\cos^{-1} x\right) = \sec \left(\tan^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)$,where $b > a > 0$,then $x =$

Which pair$(s)$ of function$(s)$ is/are equal? (where ${x}$ and $[x]$ denote the fractional part and integral part functions respectively.)

The derivative of the function $f(x) = \cos^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x - 3\sin x) \right\} + \sin^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x + 3\sin x) \right\}$ with respect to $x$ at $x = \frac{3}{4}$ is:

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

For $k \in R$,let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k$,where $0 < |x| < \frac{1}{\sqrt{2}}$,be $\alpha$ and $\beta$,where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^{2}- bx -5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$,then $\frac{b}{k^{2}}$ is equal to $......$

$\sin ^{-1}(\sin 100) + \cos ^{-1}(\cos 100) + \tan ^{-1}(\tan 100) + \cot ^{-1}(\cot 100)$ equals to: