If \(a>b>0, \sec ^{-1} \frac{a+b}{a-b}=2 \sin ^{-1} x\), then \(x\) is

If \(a>b>0, \sec ^{-1}\left(\frac{a+b}{a-b}\right)=2 \sin ^{-1} x\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)=2 \sin ^{-1} \mathrm{x}\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{1-\frac{b}{a}}{1+\frac{b}{a}}\right)=2 \sin ^{-1} x\)
\(\Rightarrow \quad \cos ^{-1}\left\{\frac{1-(\sqrt{b / a})^{2}}{1+(\sqrt{b / a})^{2}}\right\}=2 \sin ^{-1} x\)
\(\Rightarrow \quad 2 \tan ^{-1}(\sqrt{\mathrm{b} / \mathrm{a}})=2 \sin ^{-1} \mathrm{x}\)
\(\left[\because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]\)
\(\Rightarrow \quad \sin ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a+b}}\right)=\sin ^{-1} x\)
\(\left[\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}\right]\)
\(\Rightarrow \quad x=\sqrt{\frac{b}{a+b}}\)