Answer
$${\text{False}}$$
Work Step by Step
$$\eqalign{
& \sqrt {{x^2} - {a^2}} \cr
& {\text{Let }}x = a\cos \theta ,{\text{ then}} \cr
& \sqrt {{x^2} - {a^2}} = \sqrt {{{\left( {a\cos \theta } \right)}^2} - {a^2}} \cr
& {\text{Simplifying}} \cr
& \sqrt {{x^2} - {a^2}} = \sqrt {{a^2}{{\cos }^2}\theta - {a^2}} \cr
& \sqrt {{x^2} - {a^2}} = \sqrt {{a^2}\left( {{{\cos }^2}\theta - 1} \right)} \cr
& \sqrt {{x^2} - {a^2}} = \sqrt {{a^2}\left( { - {{\sin }^2}\theta } \right)} \cr
& \sqrt {{x^2} - {a^2}} = a\sqrt {\left( { - {{\sin }^2}\theta } \right)} \cr
& {\text{Therefore, the statement is false}} \cr
& {\text{Use }}x = a\cos \theta {\text{ for }}\sqrt {{a^2} - {x^2}} \cr} $$
