## Trigonometry (11th Edition) Clone

$$\frac{1-\sin^2(-\theta)}{1+\cot^2(-\theta)}=\cos^2\theta\sin^2\theta$$
$$A=\frac{1-\sin^2(-\theta)}{1+\cot^2(-\theta)}$$ $$A=\frac{1-[\sin(-\theta)]^2}{1+[\cot(-\theta)]^2}$$ - 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, $$\csc^2\theta=\frac{1}{\sin^2\theta}$$ $$A=\frac{\cos^2\theta}{\frac{1}{\sin^2\theta}}$$ $$A=\cos^2\theta\sin^2\theta$$