Solution:
Given the trigonometric equation:
[tex]tan^2\theta -sin^2\theta =sin^4\theta \:sec^2\theta [/tex]Where
[tex]\begin{gathered} tan^2\theta-sin^2\theta\Rightarrow Left\text{ hand side of the equation} \\ sin^4\theta\:sec^2\theta\Rightarrow Right\text{ hand side of the equation} \end{gathered}[/tex]Starting from the left hand side of the equation,
[tex]\begin{equation*} tan^2\theta-sin^2\theta \end{equation*}[/tex]step 1: Use a quotient identity.
[tex]\begin{gathered} tan\text{ }\theta=\frac{sin\theta}{cos\theta} \\ \Rightarrow tan^2\theta-sin^2\theta=\frac{sin^2\theta}{cos^2\theta}-sin^2\theta \end{gathered}[/tex]Step 2: Simplify the expression on the left-hand side of the equation.
Thus,
[tex]\begin{gathered} \begin{equation*} \frac{sin^2\theta}{cos^2\theta}-sin^2\theta \end{equation*} \\ =\frac{sin^2\theta-cos^2\theta sin^2\theta}{cos^2\theta} \end{gathered}[/tex]Step 3: Factor out the common term.
Thus,
[tex]\begin{gathered} \frac{s\imaginaryI n^{2}\theta- cos^{2}\theta s\imaginaryI n^{2}\theta}{cos^{2}\theta} \\ =\frac{sin^2\theta(1-cos^2\theta)}{cos^2\theta} \end{gathered}[/tex]Step 4: Use a Pythagorean identity.
According to the Pythagorean identity,
[tex]\begin{gathered} \sin^2\theta+\cos^2\theta=1 \\ \Rightarrow sin^2\theta=1-cos^2\theta \end{gathered}[/tex]thus,
[tex]\begin{gathered} \frac{s\imaginaryI n^{2}\theta- cos^{2}\theta s\imaginaryI n^{2}\theta}{cos^{2}\theta}=\frac{sin^2\theta(1-cos^2\theta)}{cos^2\theta} \\ \Rightarrow\frac{s\imaginaryI n^2\theta(sin^2\theta)}{cos^2\theta}(pythagoraen\text{ identities\rparen} \\ thus,\text{ we have} \\ \frac{sin^4\theta}{cos^2\theta} \end{gathered}[/tex]Step 5: Use a reciprocal identity.
[tex]\begin{gathered} sec\text{ }\theta=\frac{1}{cos\text{ }\theta} \\ thus, \\ \frac{s\imaginaryI n^{4}\theta}{cos^{2}\theta}=sin^4\theta\times\frac{1}{cos^2\theta}=sin^4\theta\times sec^2\theta \\ \Rightarrow sin^4\theta sec^2\theta \end{gathered}[/tex]Hence, the order in which the steps were performed are:
[tex]IV,II,I,III,V[/tex]