Answer
See the explanation below.
Work Step by Step
$\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1-{{\tan }^{2}}x}$
By using the quotient identity of trigonometry ${{\tan }^{2}}x=\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$ , the above expression can be further simplified as:
$\begin{align}
& \frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1-{{\tan }^{2}}x}=\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1-\frac{{{\sin }^{2}}x}{co{{s}^{2}}x}} \\
& =\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{co{{s}^{2}}x}}
\end{align}$
Then, rewrite the main fraction bar as $\div $:
$\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{co{{s}^{2}}x}}=\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1}\div \frac{co{{s}^{2}}x-{{\sin }^{2}}x}{co{{s}^{2}}x}$
Invert the divisor and multiply:
$\begin{align}
& \frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1}\div \frac{co{{s}^{2}}x-{{\sin }^{2}}x}{co{{s}^{2}}x}=\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1}.\frac{co{{s}^{2}}x}{co{{s}^{2}}x-{{\sin }^{2}}x} \\
& =co{{s}^{2}}x
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{co{{s}^{2}}x-{{\sin }^{2}}x}{1-{{\tan }^{2}}x}={{\cos }^{2}}x$.
