Answer
See the full explanation below.
Work Step by Step
${{\operatorname{\ln e}}^{{{\tan }^{2}}x-{{\sec }^{2}}x}}$
We use the value of $\operatorname{\ln e}$, which is equal to $1$, and take out the common “minus” sign to simplify the provided expression,
$\begin{align}
& {{\operatorname{lne}}^{{{\tan }^{2}}x-{{\sec }^{2}}x}}={{\tan }^{2}}x-{{\sec }^{2}}x \\
& =-\left( -{{\tan }^{2}}x+{{\sec }^{2}}x \right)
\end{align}$
Rearrange the terms and apply the Pythagorean identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$, which can be further derived as, ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. Then,
$\begin{align}
& -\left( -{{\tan }^{2}}x+{{\sec }^{2}}x \right)=-\left( {{\sec }^{2}}x-{{\tan }^{2}}x \right) \\
& =-1
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
${{\operatorname{\ln e}}^{{{\tan }^{2}}x-{{\sec }^{2}}x}}=-1$.
Hence, it is proved that ${{\operatorname{\ln e}}^{{{\tan }^{2}}x-{{\sec }^{2}}x}}=-1$.
