A) \[-\sqrt{\frac{b}{a+b}}\]
B) \[\sqrt{\frac{b}{a+b}}\]
C) \[-\sqrt{\frac{a}{a+b}}\]
D) \[\sqrt{\frac{a}{a+b}}\]
Correct Answer: B
Solution :
If \[a>b>0,\] \[{{\sec }^{-1}}\left( \frac{a+b}{a-b} \right)=2{{\sin }^{-1}}x\] \[\Rightarrow \] \[{{\cos }^{-1}}\left( \frac{a-b}{a+b} \right)=2{{\sin }^{-1}}x\] \[\Rightarrow \] \[{{\cos }^{-1}}\left( \frac{1-\frac{b}{a}}{1+\frac{b}{a}} \right)=2{{\sin }^{-1}}x\] \[\Rightarrow \] \[{{\cos }^{-1}}\left\{ \frac{1-{{(\sqrt{b/a})}^{2}}}{1+{{(\sqrt{b/a})}^{2}}} \right\}=2{{\sin }^{-1}}x\] \[\Rightarrow \] \[2{{\tan }^{-1}}(\sqrt{b/a})=2{{\sin }^{-1}}x\] \[\left[ \because \,\,2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]\] \[\Rightarrow \] \[{{\sin }^{-1}}\left( \frac{\sqrt{b}}{\sqrt{a+b}} \right)={{\sin }^{-1}}x\] \[\left[ \because \,\,{{\tan }^{-1}}x={{\sin }^{-1}}\frac{x}{\sqrt{1+{{x}^{2}}}} \right]\] \[\Rightarrow \] \[x=\sqrt{\frac{b}{a+b}}\]
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