Answer
$$\frac{1}{\sin\alpha-1}-\frac{1}{\sin\alpha+1}=-2\sec^2\alpha$$
Work Step by Step
$$A=\frac{1}{\sin\alpha-1}-\frac{1}{\sin\alpha+1}$$
$$A=\frac{\sin\alpha+1}{(\sin\alpha-1)(\sin\alpha+1)}-\frac{\sin\alpha-1}{(\sin\alpha-1)(\sin\alpha+1)}$$
$$A=\frac{\sin\alpha+1-(\sin\alpha-1))}{(\sin\alpha-1)(\sin\alpha+1)}$$
$$A=\frac{\sin\alpha+1-\sin\alpha+1}{(\sin\alpha-1)(\sin\alpha+1)}$$
$$A=\frac{2}{(\sin\alpha-1)(\sin\alpha+1)}$$
$$A=\frac{2}{\sin^2\alpha-1}$$ (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, $$A=-2\sec^2\alpha$$
