How do you simplify \[\dfrac{1}{{{{\cos }^2}x}} - 1\]?

Hint:We are to solve this problem involving cosine function. So, we will accordingly use different operations after taking the LCM of the cosine function and making the whole operation as one fraction. Then we will use the various trigonometric properties and identities and convert the trigonometric operation into its simplest form and find its solution. So, let us see how to solve this problem.


Complete step by step solution:
We are given the trigonometric operation is,
\[\dfrac{1}{{{{\cos }^2}x}} - 1\]
Now, taking the LCM of the operation, we get,
$ = \dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}$
Now, we know, ${\sin ^2}x + {\cos ^2}x = 1$
Subtracting ${\cos ^2}x$ from both the sides of the equation, we get,
$ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Therefore, by using this property in the given trigonometric operation, we get,
$ = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}$
We know, the tangent function is, $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
Now, taking the square over the whole operation, we get,
$ = {\left( {\dfrac{{\sin x}}{{\cos x}}} \right)^2}$
Therefore, using the property of tangent function, we get,
$ = {\tan ^2}x$
Therefore, the simplified form of the trigonometric operation \[\dfrac{1}{{{{\cos }^2}x}} - 1\] is ${\tan ^2}x$.

Note:
There is also an alternative way to solve this problem. We know, secant function is the reciprocal of cosine function, that is, $\sec x = \dfrac{1}{{\cos x}}$. By using this property we can convert the whole operation in terms of secant function. We know that, \[{\tan ^2}x + 1 = {\sec ^2}x\]. From which we can convert the identity into \[{\tan ^2}x = {\sec ^2}x - 1\]. Then by substituting this property we can get our required result that is ${\tan ^2}x$.