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