Trigonometry (11th Edition) Clone

$$\frac{1}{\sin\alpha-1}-\frac{1}{\sin\alpha+1}=-2\sec^2\alpha$$
$$A=\frac{1}{\sin\alpha-1}-\frac{1}{\sin\alpha+1}$$ $$A=\frac{\sin\alpha+1}{(\sin\alpha-1)(\sin\alpha+1)}-\frac{\sin\alpha-1}{(\sin\alpha-1)(\sin\alpha+1)}$$ $$A=\frac{\sin\alpha+1-(\sin\alpha-1))}{(\sin\alpha-1)(\sin\alpha+1)}$$ $$A=\frac{\sin\alpha+1-\sin\alpha+1}{(\sin\alpha-1)(\sin\alpha+1)}$$ $$A=\frac{2}{(\sin\alpha-1)(\sin\alpha+1)}$$ $$A=\frac{2}{\sin^2\alpha-1}$$ (for $a^2-b^2=(a-b)(a+b)$) - Pythagorean Identity: $$\cos^2\alpha=1-\sin^2\alpha$$ $$-\cos^2\alpha=\sin^2\alpha-1$$ Therefore, $$A=\frac{2}{-\cos^2\alpha}$$ $$A=-2(\frac{1}{\cos\alpha})^2$$ - Reciprocal Identity: $$\sec\alpha=\frac{1}{\cos\alpha}$$ So, $$A=-2\sec^2\alpha$$