I'm doing practice problems out of Trigonometry 10th ed. Hornsby and ran into a question.
Section 5.1 question 71:
Write each expression in terms of sine and cosine, and simplify so that no quotients appear in the final expression and all functions are of $\theta$.
$\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)}$
The book has the answer $\sin^2 \theta \cos^2 \theta$. However, I cannot figure out how to get to this answer.
I started by pulling out the negative from $\theta$:
$\frac{1+\cos^2\theta}{1-\tan^2\theta}$
Then I changed tangent into sine and cosine: $\frac{1+\cos^2\theta}{1-\frac{\sin\theta}{\cos\theta}}$
Then multiplied by the reciprocal:
$(1+\cos^2\theta)(1-\frac{\cos\theta}{\sin\theta})$
And this is where things got really confusing as I had no idea what I could do with the result of the above expression.