$\text{ Range of the function }f\left(x\right)=\sqrt{{\cos }^{-1}\left(\sqrt{{\log }_{4}x}\right)-\frac{\pi }{2}}$$+{\sin }^{-1}\left(\frac{1+x^2}{4x}\right)\text{ is }$

$f\left(x\right)=\sqrt{-{\sin }^{-1}\left(\sqrt{\log {}_{4}x}\right)}+{\sin }^{-1}\left(\frac{1+x^2}{4x}\right)$     $\left(\sin ce {\sin }^{-1}x + {\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right)$

       $domain of first function is \left\{1\right\} only$ $$\begin{gathered} \left\{\sin ce -{\sin }^{-1 }\left(\sqrt{\log {}_{4}x}\right)\ge 0 \\ \Rightarrow {\sin }^{-1}\left(\sqrt{\log {}_{4}x}\right)\le 0 \\ \Rightarrow \sqrt{\log {}_{4}x}\le \sin 0 \\ \Rightarrow \log {}_{4}x =0 \\ \Rightarrow x=1 \right\} \end{gathered}$$

   $therefore f\left(x\right) domain also \left\{1\right\}$      $\left( \sin ce Domain of \left(f+g\right)=\left(Domain of f\right)\cap \left(Domain of g\right)\right)$

 $$\begin{gathered} Range of f\left(x\right) = \left\{f\left(1\right)\right\} =\left\{\sqrt{-{\sin }^{-1 }0} +{\sin }^{-1}\left(\frac{1+1}{4}\right)\right\} \\ =\left\{{\sin }^{-1}\frac{1}{2}\right\} \\ =\left\{\frac{\mathrm{\pi }}{6}\right\} \end{gathered}$$