If a b 0 and sec^{-1} left( frac{a+b}{a-b | vedclass

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(B) Given $\sec^{-1} \left( \frac{a+b}{a-b} \right) = 2 \sin^{-1} x$. Taking the reciprocal,we get $\cos^{-1} \left( \frac{a-b}{a+b} \right) = 2 \sin^

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$\sqrt{\frac{b}{a+b}}$

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(B) Given $\sec^{-1} \left( \frac{a+b}{a-b} ight) = 2 \sin^{-1} x$. Taking the reciprocal,we get $\cos^{-1} \left( \frac{a-b}{a+b} ight) = 2 \sin^{-1} x$. Dividing the numerator and denominator by $a$,we have $\cos^{-1} \left( \frac{1 - b/a}{1 + b/a} ight) = 2 \sin^{-1} x$. Using the identity $\cos^{-1} \left( \frac{1 - y^2}{1 + y^2} ight) = 2 	an^{-1} y$,where $y = \sqrt{b/a}$,we get $2 	an^{-1} \left( \sqrt{\frac{b}{a}} ight) = 2 \sin^{-1} x$. Dividing by $2$,we have $	an^{-1} \left( \sqrt{\frac{b}{a}} ight) = \sin^{-1} x$. Using the identity $	an^{-1} y = \sin^{-1} \left( \frac{y}{\sqrt{1+y^2}} ight)$,we get $\sin^{-1} \left( \frac{\sqrt{b/a}}{\sqrt{1 + b/a}} ight) = \sin^{-1} x$. Simplifying the expression inside,we get $\sin^{-1} \left( \sqrt{\frac{b/a}{(a+b)/a}} ight) = \sin^{-1} x$. Thus,$x = \sqrt{\frac{b}{a+b}}$.

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

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