Verified Answer :

Hippalectryon

\(f(x)=\sqrt{x^2-1},g(x)=\sqrt{x-1}\)

Hence \(f(x)/g(x)=\sqrt{\frac{x^2-1}{x-1}}=\sqrt{\frac{(x+1)(x-1)}{x-1}}=\sqrt{x+1}\)

Panoyin

\(f(x) = \sqrt{x^{2} - 1}\)
\(g(x) = \sqrt{x - 1}\)

\(\frac{f(x)}{g(x)} = \sqrt{\frac{x^{2} - 1}{x - 1}}\)

\(\frac{f(x)}{g(x)} = \frac{\sqrt{x^{2} - 1}}{\sqrt{x - 1}}\)

\(\frac{f(x)}{g(x)} = \frac{\sqrt{x^{2} + x - x - 1}}{\sqrt{x - 1}}\)

\(\frac{f(x)}{g(x)} = \frac{\sqrt{(x - 1)(x + 1)}}{\sqrt{x - 1}}\)

\(\frac{f(x)}{g(x)} = \sqrt{x + 1}\)

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