Given:
[tex]\cos ^4x\tan ^2x[/tex]To rewrite: The expression below in terms containing only the first powers of cosine.
Explanation:
Using the identity,
[tex]\tan ^2x=\frac{\sin ^2x}{\cos ^2x}[/tex]We can write,
[tex]\begin{gathered} \cos ^4x\tan ^2x=\cos ^4x\cdot\frac{\sin ^2x}{\cos ^2x} \\ =\cos ^2x\cdot\sin ^2x \end{gathered}[/tex]Using the identity,
[tex]\sin ^2x=1-\cos ^2x[/tex]So, we can write,
[tex]\begin{gathered} =\cos ^2x(1-\cos ^2x) \\ =\cos ^2x-\cos ^4x \\ =\cos ^2x-(\cos ^2x)^2 \end{gathered}[/tex]Using the identity,
[tex]\cos ^2x=\frac{1+\cos2x}{2}[/tex]Substituting we get,
[tex]\begin{gathered} =\frac{1+\cos2x}{2}-(\frac{1+\cos2x}{2})^2 \\ =\frac{1+\cos2x}{2}-(\frac{1+\cos ^22x+2\cos 2x}{4}) \\ =\frac{1+\cos2x}{2}-(\frac{1+\cos^22x+2\cos2x}{4}) \\ =\frac{1}{2}+\frac{\cos2x}{2}-\frac{1}{4}-\frac{\cos^22x}{4}-\frac{\cos 2x}{2} \\ =\frac{1}{4}-\frac{\cos^22x}{4} \\ =\frac{1}{4}-\frac{\frac{1+\cos4x}{2}}{4} \\ =\frac{1}{4}-\frac{1+\cos 4x}{8} \\ =\frac{1}{4}-\frac{1}{8}-\frac{\cos 4x}{8} \\ =\frac{1}{8}-\frac{\cos4x}{8} \\ =\frac{1}{8}(1-cos4x) \end{gathered}[/tex]Final answer: The power reduced form in terms of only first powers of cosine is,
[tex]\frac{1}{8}(1-\cos 4x)[/tex]