How do you simplify the expression \[1 - {\sec ^2}x\]?

Hint: In this question the trigonometric expression is given, for simplify this expression we will use the basic trigonometric identity. And there are some trigonometric identities as given below.
\[ \Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x\]
\[ \Rightarrow \sin x = \dfrac{1}{{\csc x}}\]
\[ \Rightarrow \cos x = \dfrac{1}{{\sec x}}\]
\[ \Rightarrow \tan x = \dfrac{1}{{\cot x}}\]

Complete step by step answer:
In this question, the word trigonometric identity is used. First, we know about trigonometric identities.
The trigonometric identities are defined as the equations that are true for right-angle triangles.
We determine the trigonometric identities by using the right-angle triangle. Its angle is \[x\]. Here $b$ is the length of the base, $p$ is the height and $h$ is the length of the hypotenuse.
There are three main functions in trigonometry: sin, cos, and tan.
Then we write these functions according to the right angle triangle.
\[
  \sin x = \dfrac{p}{h} \\
  \cos x = \dfrac{b}{h} \\
  \tan x = \dfrac{p}{b} \\
 \]
Then,
\[\tan x\] is written as below.
\[ \Rightarrow \dfrac{{\sin x}}{{\cos x}} = \dfrac{{p/h}}{{b/h}} = \dfrac{p}{b} = \tan x\]
Then,
\[ \Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}}\]
And another trigonometric function is written as below (by using right angle triangle).
\[
  \csc x = \dfrac{h}{p} = \dfrac{1}{{\sin x}} \\
  \sec x = \dfrac{h}{b} = \dfrac{1}{{\cos x}} \\
  \cot x = \dfrac{b}{p} = \dfrac{1}{{\tan x}} \\
 \]

We can also write it as,
\[
  \sin x = \dfrac{p}{h} = \dfrac{1}{{\cos ecx}} \\
  \cos x = \dfrac{b}{h} = \dfrac{1}{{\sec x}} \\
  \tan x = \dfrac{p}{b} = \dfrac{1}{{\cot x}} \\
 \]

Now, in the question, the trigonometric form is given which is simplified as below.
\[1 - {\sec ^2}x\]
We know that according to the right angle triangle \[\sec x = \dfrac{h}{b}\].
Put this value in the above equation.
Then, the above equation is written as.
\[ \Rightarrow 1 - \dfrac{{{h^2}}}{{{b^2}}}\]
Now, we simplify the above equation.
Then,
\[ \Rightarrow \dfrac{{{b^2} - {h^2}}}{{{b^2}}}\]
We know that, according to Pythagoras theorem.
\[
   \Rightarrow {h^2} = {p^2} + {b^2} \\
   \Rightarrow {b^2} - {h^2} = - {p^2} \\
 \]
Now, we put the \[{b^2} - {h^2}\] value in the above equation.
Then,
\[ \Rightarrow - \dfrac{{{p^2}}}{{{b^2}}}\]
We know that \[\dfrac{p}{b} = \tan x\]. Put that value in above.
Then,
\[\therefore - {\tan ^2}x\]
Therefore, \[1 - {\sec ^2}x\] is simplified as \[ - {\tan ^2}x\].

Note: If we want to find the result of trigonometric identities, then you are suggested to use the right-angle triangle and Pythagoras theorem. By using these two, we can easily find the result of any trigonometric identity.