How do you simplify ${{\left( \sec \left( x \right) \right)}^{2}}-1$?

Hint: While simplifying a trigonometric expression, we should always aim to reduce the expression to have the least number of functions and variables possible. We can use trigonometric identities to simplify trigonometric expressions like these.

Complete Step by Step Solution:
The given trigonometric expression is ${{\left( \sec \left( x \right) \right)}^{2}}-1$. Here we can see that it has a trigonometric function and a constant. Let us try to eliminate the constant.
Let us consider the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. If we divide this identity by ${{\cos }^{2}}x$, we get
$\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}+\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$
In the above equation, we know that $\dfrac{\sin x}{\cos x}=\tan x$
$\Rightarrow {{\tan }^{2}}x+1=\dfrac{1}{{{\cos }^{2}}x}$
Also, cosine is an inverse function of secant. Therefore
$\Rightarrow {{\tan }^{2}}x+1={{\sec }^{2}}x$
We can rewrite the above equation as
$\Rightarrow {{\sec }^{2}}x-1={{\tan }^{2}}x$

Hence the simplified form of ${{\left( \sec \left( x \right) \right)}^{2}}-1$ is equal to ${{\tan }^{2}}x$.

Note:
We should be familiar with the trigonometric identities and the relationship between various trigonometric ratios to simplify these types of expressions. For example, in this problem, we have used the relations $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{1}{\cos x}=\sec x$ in order to simplify the expression.