Rewrite $\sec^2 (x) \cdot \sin^2 (x) $in terms of $\cos (x)$

Question

Write $\sec^2 (x) \cdot \sin^2 (x)$ in terms of $\cos (x)$.

I was not able to answer this question on a pre-calculus exam and my professor will not explain to me how to answer it; I wish to learn.

Here is what I did on the test:

$\sec^2 (x) \cdot \sin^2 (x)$ = $\frac{\sqrt{1}}{\sqrt{\cos^2 (x)}} \cdot \ \sqrt{1-\cos^2 (x)} = \frac{1}{\cos (x)} \cdot \frac{\pm\sqrt{1-\cos^2}}{1} = \frac{\pm\sqrt{1-\cos^2}}{\cos (x)}$

What did I do wrong here?

Thank you in advance.

Answer 1

$\sec x = \frac{1}{\cos x} \\ \sin^2x = 1 - \cos^2x$

Put it all together.

Answer 2

Notice that we have that: $$\sec^2 x = \frac{1}{\cos^2 x} $$ $$ \sin^2x = 1 - \cos^2x$$

So if we multiply the two and use distributivity: $$(\sec^2 x) \cdot(\sin^2x )=(\frac{1}{\cos^2 x}) \cdot(1 - \cos^2x)=\frac{1}{\cos^2 x} -\frac{\cos^2 x}{\cos^2 x}= \frac{1}{\cos^2 x} -1$$

Notice now that we expressed this in terms of $\cos(x)$, we now have a function in terms of $\cos(x)$, namely: $$ f(\cos x )=\frac{1}{\cos^2 x} -1$$ Or if we let $z=\cos(x)$ $$ f(z )=\frac{1}{z^2} -1$$ would you not say that this is expressed in terms of $z$ $(=\cos(x))$?

Answer 3

$sec^2(x)\bullet sin^2(x)={1\over{cos^2(x)}}\left(1-cos^2(x)\right)={1\over{cos^2(x)}}-1$