If cot(cos^{-1} x) = sec(tan^{-1} frac{a}{sqr | vedclass

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(C) Let $\cos^{-1} x = 	heta$, then $\cos 	heta = x$. Thus, $\cot(\cos^{-1} x) = \cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{x}{\sqrt{1 -

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

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(C) Let $\cos^{-1} x = 	heta$, then $\cos 	heta = x$. Thus, $\cot(\cos^{-1} x) = \cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{x}{\sqrt{1 - x^2}}$.Let $	an^{-1} \frac{a}{\sqrt{b^2 - a^2}} = \phi$, then $	an \phi = \frac{a}{\sqrt{b^2 - a^2}}$.Using the identity $\sec^2 \phi = 1 + 	an^2 \phi$, we get $\sec^2 \phi = 1 + \frac{a^2}{b^2 - a^2} = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2}$.Therefore, $\sec \phi = \frac{b}{\sqrt{b^2 - a^2}}$.Equating the two sides: $\frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}$.Squaring both sides: $\frac{x^2}{1 - x^2} = \frac{b^2}{b^2 - a^2}$.$x^2(b^2 - a^2) = b^2(1 - x^2) = b^2 - b^2 x^2$.$x^2(b^2 - a^2 + b^2) = b^2 \implies x^2(2b^2 - a^2) = b^2$.$x^2 = \frac{b^2}{2b^2 - a^2} \implies x = \frac{b}{\sqrt{2b^2 - a^2}}$.

If $\cot(\cos^{-1} x) = \sec(\tan^{-1} \frac{a}{\sqrt{b^2 - a^2}})$, then the value of $x$ is

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