If f(x)= sin

$f(x)=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x}}{1+2^{2 x}}\right)\right)$ at $x=1 ; 2^{2 x}=4$

for $\sin \left(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right)$

Let $\tan ^{-1} \mathrm{x}=\theta ; \quad \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

$\therefore \quad \sin \left(\cos ^{-1} \cos 2 \theta\right)=\sin 2 \theta$

If $x>1 \Rightarrow \frac{\pi}{3}>\theta>\frac{\pi}{4}$,

$\left\{\therefore \quad \pi>2 \theta>\frac{\pi}{2}\right.$\}

$=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$\therefore \quad f^{1}(1)=\frac{20 \ln 2-32 \ln 2}{25}=-\frac{12}{25} \ln 2$

So, $a=25, b=12 \Rightarrow\left|a^{2}-b^{2}\right|=25^{2}-12^{2}$

$=625-144$

$=481$