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

Hint: To solve this problem we have to know basic trigonometric equations. First we have to take the identity which has a relation between numerator and denominator. From there we can apply some basic mathematical formulas so that we arrive at the solution.

Complete step-by-step answer:
First we have to know the trigonometric identities
\[\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & 1+{{\tan }^{2}}x={{\sec }^{2}}x \\
 & 1+{{\cot }^{2}}x={{\csc }^{2}}x \\
\end{align}\]
From that we can take \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
So we can rewrite the equation as
\[{cos}^{2}x=1-{sin}^{2}x………(1)\]
Using \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
We can write
\[\Rightarrow 1-{{\sin }^{2}}x=\left( 1-\sin x \right)\left( 1+\sin x \right).......\left( 2 \right)\]
Since \[1-{{\sin }^{2}}x\] can also be written as \[{{1}^{2}}-{{\sin }^{2}}x\]
So, From above equations 1 & 2 we can write
\[\Rightarrow {{\cos }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right)\]
\[\dfrac{{{\cos }^{2}}x}{1-\sin x}=1+\sin x\]
We can do it in this way also
\[\Rightarrow {{\cos }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right)\]
Now we are dividing the equation with \[1-\sin x\] on both sides
\[\Rightarrow \dfrac{{{\cos }^{2}}x}{1-\sin x}=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{1-\sin x}\]
Now we proceed to cancel \[1-\sin x\] in both numerator and denominator because they configure a so called avoidable discontinuity.
Avoidable discontinuity:
An avoidable discontinuity in a point \[x=a\] occurs when the side limits coincide, but the value of the function at this point is not, that is to say:
\[\displaystyle \lim_{x\to {{a}^{-}}}f(x)=\displaystyle \lim_{x\to {{a}^{+}}}=Lf{{(a)}^{{}}}\ne L\]
It is reasonable that we call this type of discontinuity avoidable since the function in the discontinuity point seems continuous, but the point in particular does not satisfy the definition of continuity, so only by adding this point to the definition of the function, can we obtain a continuous function.
We will get
\[\dfrac{{{\cos }^{2}}x}{1-\sin x}=1+\sin x\]
From this we can identify \[\dfrac{{{\cos }^{2}}x}{1-\sin x}\] as \[1+\sin x\] .

Note: We can solve this in another way using trigonometric identities directly. While solving the question students may complicate the problem using advanced trigonometric identities but with only basic identities and simple maths formulas we can solve the question easily.