If cot left(cos^{-1} xright) = sec left(tan^{ | vedclass

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(A) Let $	heta = 	an^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}ight)$. Then $	an 	heta = \frac{a}{\sqrt{b^2-a^2}}$.Using the identity $\sec^2 	heta =

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$\frac{b}{\sqrt{2b^2-a^2}}$

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(A) Let $	heta = 	an^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}ight)$. Then $	an 	heta = \frac{a}{\sqrt{b^2-a^2}}$.Using the identity $\sec^2 	heta = 1 + 	an^2 	heta$,we have $\sec^2 	heta = 1 + \frac{a^2}{b^2-a^2} = \frac{b^2-a^2+a^2}{b^2-a^2} = \frac{b^2}{b^2-a^2}$.Thus,$\sec 	heta = \frac{b}{\sqrt{b^2-a^2}}$.Now,let $\phi = \cos^{-1} x$. Then $\cos \phi = x$,which implies $\sin \phi = \sqrt{1-x^2}$.So,$\cot \phi = \frac{\cos \phi}{\sin \phi} = \frac{x}{\sqrt{1-x^2}}$.The given equation is $\cot \phi = \sec 	heta$,so $\frac{x}{\sqrt{1-x^2}} = \frac{b}{\sqrt{b^2-a^2}}$.Squaring both sides: $\frac{x^2}{1-x^2} = \frac{b^2}{b^2-a^2}$.Cross-multiplying: $x^2(b^2-a^2) = b^2(1-x^2) = b^2 - b^2x^2$.$x^2(b^2-a^2+b^2) = b^2$.$x^2(2b^2-a^2) = b^2$.$x^2 = \frac{b^2}{2b^2-a^2}$.Taking the square root,$x = \frac{b}{\sqrt{2b^2-a^2}}$.

If $\cot \left(\cos^{-1} x\right) = \sec \left(\tan^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)$,where $b > a > 0$,then $x =$

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