## Precalculus: Mathematics for Calculus, 7th Edition

$\sqrt{1+\Big(\dfrac{x}{\sqrt{1-x^{2}}}\Big)^{2}}=\sqrt{\dfrac{1}{1-x^{2}}}$
$\sqrt{1+\Big(\dfrac{x}{\sqrt{1-x^{2}}}\Big)^{2}}$ Evaluate the power inside the root: $\sqrt{1+\Big(\dfrac{x}{\sqrt{1-x^{2}}}\Big)^{2}}=\sqrt{1+\dfrac{x^{2}}{1-x^{2}}}=...$ Evaluate the sum of fractions inside the root: $...=\sqrt{\dfrac{1-x^{2}+x^{2}}{1-x^{2}}}=\sqrt{\dfrac{1}{1-x^{2}}}$