Answer
$$\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)}=\sin^2\theta\cos^2\theta$$
Work Step by Step
$$A=\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)}$$
$$A=\frac{1-[\cos(-\theta)]^2}{1+[\tan(-\theta)]^2}$$
- Negative-angle Identity:
$$\cos(-\theta)=\cos\theta$$
$$\tan(-\theta)=-\tan\theta$$
Therefore, $[\cos(-\theta)]^2=[\cos\theta]^2=\cos^2\theta$ and $[\tan(-\theta)]^2=[-\tan\theta]^2=\tan^2\theta$ (since $[-A]^2=A^2$)
Replace into $A$:
$$A=\frac{1-\cos^2\theta}{1+\tan^2\theta}$$
- Pythagorean Identity:
$$1-\cos^2\theta=\sin^2\theta$$
$$1+\tan^2\theta=\sec^2\theta$$
Replace into $A$:
$$A=\frac{\sin^2\theta}{\sec^2\theta}$$
- Reciprocal Identity:
$$\sec\theta=\frac{1}{\cos\theta}$$
Then, $$\sec^2\theta=\frac{1}{\cos^2\theta}$$
$$A=\frac{\sin^2\theta}{\frac{1}{\cos^2\theta}}$$
$$A=\sin^2\theta\cos^2\theta$$
