How do you express $\cos \theta -{{\cos }^{2}}\theta +\sec \theta $ in terms of $\sin \theta $?

Hint: We first try to establish the trigonometric identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. We find the value of \[{{\cos }^{2}}\theta \] in the form of \[\sin \theta \] where \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \]. We also have \[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\]. We only simplify the $\cos \theta +\sec \theta $ part and put the values to find $\cos \theta -{{\cos }^{2}}\theta +\sec \theta $ in terms of $\sin \theta $.

Complete step-by-step answer:
We have the identity theorem of trigonometric values where \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
From the relation we get \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \].
We take square root value on both sides to get
\[\begin{align}
  & \sqrt{{{\cos }^{2}}\theta }=\sqrt{1-{{\sin }^{2}}\theta } \\
 & \Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta } \\
\end{align}\]
We first try to simplify the equation $\cos \theta -{{\cos }^{2}}\theta +\sec \theta $.
We keep the term $-{{\cos }^{2}}\theta $ in $\cos \theta -{{\cos }^{2}}\theta +\sec \theta $ as it is.
We simplify the rest of the expression $\cos \theta +\sec \theta $.
We know the relation where $\sec \theta =\dfrac{1}{\cos \theta }$.
We put the value and get $\cos \theta +\sec \theta =\cos \theta +\dfrac{1}{\cos \theta }$.
We simplify the equation to get $\cos \theta +\dfrac{1}{\cos \theta }=\dfrac{{{\cos }^{2}}\theta +1}{\cos \theta }$.
Now the final expression becomes
$\cos \theta -{{\cos }^{2}}\theta +\sec \theta =\dfrac{{{\cos }^{2}}\theta +1}{\cos \theta }-{{\cos }^{2}}\theta $.
Now we place the values of \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \] and \[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\] to get
$\begin{align}
  & \cos \theta -{{\cos }^{2}}\theta +\sec \theta \\
 & =\dfrac{{{\cos }^{2}}\theta +1}{\cos \theta }-{{\cos }^{2}}\theta \\
 & =\dfrac{1-{{\sin }^{2}}\theta +1}{\sqrt{1-{{\sin }^{2}}\theta }}-\left( 1-{{\sin }^{2}}\theta \right) \\
\end{align}$
Now we simplify the equation to get
$\begin{align}
  & \cos \theta -{{\cos }^{2}}\theta +\sec \theta \\
 & =\dfrac{1-{{\sin }^{2}}\theta +1}{\sqrt{1-{{\sin }^{2}}\theta }}-\left( 1-{{\sin }^{2}}\theta \right) \\
 & ={{\sin }^{2}}\theta +\dfrac{2-{{\sin }^{2}}\theta }{\sqrt{1-{{\sin }^{2}}\theta }}-1 \\
\end{align}$
The expression of $\cos \theta -{{\cos }^{2}}\theta +\sec \theta $ in terms of $\sin \theta $ is ${{\sin }^{2}}\theta +\dfrac{2-{{\sin }^{2}}\theta }{\sqrt{1-{{\sin }^{2}}\theta }}-1$.

Note: The square root portion can’t be simplified further. We keep it as it is. For the simplified form we didn’t take the negative version. The modulus of the function omits the possibility.