The number of solution of the equation ${\cos }^{-1} \left(\frac{1+{\mathrm{x}}^2}{2\mathrm{x}}\right)-{\cos }^{-1} \mathrm{x}=\frac{\mathrm{\pi }}{2}+{\sin }^{-1} \mathrm{x}$ is

$\begin{array}{rr}From the given equation {\cos }^{-1} \left(\frac{1+{\mathrm{x}}^2}{2\mathrm{x}}\right)=\frac{\mathrm{\pi }}{2}+\left({\sin }^{-1} \mathrm{x}+{\cos }^{-1} \mathrm{x}\right)& \\ \Rightarrow {\cos }^{-1} \left(\frac{1+{\mathrm{x}}^2}{2\mathrm{x}}\right)=\frac{\mathrm{\pi }}{2}& +\frac{\mathrm{\pi }}{2}\\ \Rightarrow {\cos }^{-1} \left(\frac{1+{\mathrm{x}}^2}{2\mathrm{x}}\right)=\mathrm{\pi }& \mathrm{ }\\ \Rightarrow \frac{1+{\mathrm{x}}^2}{2\mathrm{x}}=\cos \pi & \\ \Rightarrow \frac{1+{\mathrm{x}}^2}{2\mathrm{x}}=-1& \\ \Rightarrow 1+x^2=-2x& \\ \Rightarrow 1+x^2+2x=0& \\ \Rightarrow {(1+x)}^2=0& \\ \Rightarrow 1+x=0& \\ \Rightarrow x=-1& \end{array}$