Answer
$\sqrt{1+\Big(\dfrac{x}{\sqrt{1-x^{2}}}\Big)^{2}}=\sqrt{\dfrac{1}{1-x^{2}}}$
Work Step by Step
$\sqrt{1+\Big(\dfrac{x}{\sqrt{1-x^{2}}}\Big)^{2}}$
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:
$...=\sqrt{\dfrac{1-x^{2}+x^{2}}{1-x^{2}}}=\sqrt{\dfrac{1}{1-x^{2}}}$
