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

Hint: In this question we have simplified the given trigonometric form. Next, we use some trigonometric identities and then simplify to arrive at our final answer. Next, we rearrange the trigonometric functions .And also we are going to division and LCM in complete step by step solution.
The trigonometric functions (also called circular functions, angle functions or goniometric functions) are real functions which relate an angle of a right-angled triangle to ratios of two side lengths.

Complete step-by-step solution:
It is given in the question stated as,
\[ \Rightarrow \dfrac{{{{\sec }^2}x - 1}}{{{{\sin }^2}x}}\]
First, convert all of the trigonometric functions to$\sin \left( x \right)$and$\cos \left( x \right)$:
Now, we change${\sec ^2}x$.
Put ${\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}$
We get,
\[ \Rightarrow \dfrac{{\dfrac{1}{{{{\cos }^2}(x)}} - 1}}{{{{\sin }^2}(x)}}\]
Next, we take the LCM of numerator in the above term and we get
\[ \Rightarrow \dfrac{{\dfrac{{1 - {{\cos }^2}(x)}}{{{{\cos }^2}(x)}}}}{{{{\sin }^2}(x)}}\]
Now, we use the identity\[{\sin ^2}(x) + {\cos ^2}(x) = 1\].
Then, we rearrange the above identity and put in the above numerator and we get
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^2}(x)}}{{{{\cos }^2}(x)}}}}{{{{\sin }^2}(x)}}\] . Here, \[{\sin ^2}(x) = 1 - {\cos ^2}(x)\]
\[ \Rightarrow \dfrac{{{{\sin }^2}(x)}}{{{{\cos }^2}(x) \times {{\sin }^2}(x)}}\]
Next, cancelling out the \[si{n^2}(x)\]present in both the numerator and the denominator:
We get,
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}(x)}}\]
$ \Rightarrow {\sec ^2}\left( x \right)$.
Here, \[\dfrac{1}{{{{\cos }^2}(x)}} = {\sec ^2}x\]
This is the required answer for the given trigonometric form.

The answer is ${\sec ^2}\left( x \right)$.

Note: We remember that the trigonometric functions most widely used in modern mathematics are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used. Each of these six trigonometric functions has a corresponding inverse function (called inverse trigonometric function), and an equivalent in the hyperbolic functions as well.
Sine Function : \[\sin \left( \theta \right) = \dfrac{{Opposite}}{{Hypotenuse}}\]
Cosine Function : \[\cos \left( \theta \right) = \dfrac{{Adjacent}}{{Hypotenuse}}{\text{ }}\]
Tangent Function: \[\tan \left( \theta \right) = \dfrac{{Opposite}}{{Adjacent}}{\text{ }}\]
In a right-angled triangle, the sum of the two acute angles is a right angle, that is, \[{90^ \circ }\]or $\dfrac{\pi }{2}$radians.
There is another little hard way to find the answer.
We know that the identity,
\[ \Rightarrow {\sec ^2}x - 1 = {\tan ^2}x\]
Therefore, we put the identity in given term and we get
\[ \Rightarrow \dfrac{{{{\sec }^2}\left( x \right) - 1}}{{{{\sin }^2}(x)}}\]
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( x \right)}}{{{{\sin }^2}(x)}}\]
Already we know, $\tan x = \dfrac{{\sin x}}{{\cos x}}$
Now, substitute the trigonometry identity in numerator and we get
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^2}\left( x \right)}}{{{{\cos }^2}\left( x \right)}}}}{{{{\sin }^2}(x)}}\]
\[ \Rightarrow \dfrac{{\operatorname{si} {{{n}}^2}\left( x \right)}}{{{{\cos }^2}\left( x \right)}} \times \dfrac{1}{{\operatorname{si} {{{n}}^2}(x)}}\]
Next, cancelling ${\sin ^2}\left( x \right)$and we get
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}(x)}}\]
$ \Rightarrow {\sec ^2}\left( x \right)$.
Here, \[\dfrac{1}{{{{\cos }^2}(x)}} = {\sec ^2}x\]
This is the required answer for the given trigonometric form.
The answer is ${\sec ^2}\left( x \right)$.