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(A) Let $	heta = 	ext{cosec}^{-1}x$. Then $	ext{cosec}	heta = x$,where $|x| \ge 1$ and $	heta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.We

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$	ext{cosec}(\sec^{-1}x)$

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(A) Let $	heta = 	ext{cosec}^{-1}x$. Then $	ext{cosec}	heta = x$,where $|x| \ge 1$ and $	heta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.We need to find $\sec(	ext{cosec}^{-1}x) = \sec	heta$.Since $	ext{cosec}	heta = x$,we have $\sin	heta = \frac{1}{x}$.Using the identity $\cos^2	heta = 1 - \sin^2	heta = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$,we get $\cos	heta = \frac{\sqrt{x^2 - 1}}{|x|}$.Therefore,$\sec	heta = \frac{1}{\cos	heta} = \frac{|x|}{\sqrt{x^2 - 1}}$.Now,consider $	ext{cosec}(\sec^{-1}x)$. Let $\phi = \sec^{-1}x$. Then $\sec\phi = x$,so $\cos\phi = \frac{1}{x}$.Then $\sin\phi = \sqrt{1 - \cos^2\phi} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|}$.Thus,$	ext{cosec}\phi = \frac{1}{\sin\phi} = \frac{|x|}{\sqrt{x^2 - 1}}$.Since both expressions equal $\frac{|x|}{\sqrt{x^2 - 1}}$,we have $\sec(	ext{cosec}^{-1}x) = 	ext{cosec}(\sec^{-1}x)$.

$\sec(\text{cosec}^{-1}x)$ is equal to

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