Let $f\left(x\right)={\left({\sin }^{-1}x\right)}^2-{\left({\cos }^{-1}x\right)}^2.$ If range of $f$ equals $[\frac{a{\pi }^2}{4},\frac{b{\pi }^2}{4}]$ where $a,b\in Z,$ then the value of $b-a$ is

Clearly, domain is $[-1,1]$
$f\left(x\right)={\left({\sin }^{-1}x\right)}^2-{\left({\cos }^{-1}x\right)}^2$
$=({\sin }^{-1}x+{\cos }^{-1}x)({\sin }^{-1}x-{\cos }^{-1}x)$
$=\frac{\pi }{2}({\sin }^{-1}x-(\frac{\pi }{2}-{\sin }^{-1}x))$
$\therefore f\left(x\right)=\frac{\pi }{2}(2{\sin }^{-1}x-\frac{\pi }{2})$ for all $x\in [-1,1]$

$f^{\prime }\left(x\right)=\frac{\pi }{\sqrt{1-x^2}}>0$
$f$ is a strictly increasing function.
So, $f_{max}=f\left(1\right)=\frac{\pi }{2}(2\left(\frac{\pi }{2}\right)-\frac{\pi }{2})=\frac{{\pi }^2}{4}$
and $f_{min}=f(-1)=\frac{\pi }{2}(-2\left(\frac{\pi }{2}\right)-\frac{\pi }{2})=\frac{-3{\pi }^2}{4}$
Range of $f$ is $[\frac{-3{\pi }^2}{4},\frac{{\pi }^2}{4}]\equiv [\frac{a{\pi }^2}{4},\frac{b{\pi }^2}{4}]$ (Given)
$\therefore a=-3$ and $b=1$
Hence, $b-a=1-(-3)=4$