Answer
See the explanation below.
Work Step by Step
$1-{{\tan }^{2}}x=\frac{\cos 2x}{{{\cos }^{2}}x}$
Consider the left side of the given expression and apply the quotient Identity formula.
$\begin{align}
& 1-{{\tan }^{2}}x=1-\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x} \\
& =\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x}
\end{align}$
Apply the double angle formula,
$1-{{\tan }^{2}}x=\frac{\cos 2x}{{{\cos }^{2}}x}$
Hence, it is proved that the given identity $1-{{\tan }^{2}}x=\frac{\cos 2x}{{{\cos }^{2}}x}$ holds true.
