If $\cot(\cos^{-1} x) = \sec\left(\tan^{-1} \dfrac{a}{\sqrt{b^2 - a^2}}\right)$, then the value of $x$ is

If $g(x) = x^2 + x - 2$ and $\dfrac{1}{2}(g \circ f)(x) = 2x^2 - 5x + 2$ then $f(x)$ is equal to

Let the function f be defined by $f(x) = \dfrac{x - |x|}{x}$, for $x \neq 0$ and $f(0) = 2$ then f is

The derivative of $\tan^{-1}\left(\dfrac{\sqrt{1+x^2} - 1}{x}\right)$ with respect to $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$ is...