1. **State the problem:** Simplify the expression $$\frac{1}{\sin x} + \frac{1}{\cos x} \over \frac{1}{\sin x} - \frac{1}{\cos x}$$. 2. **Write the expression clearly:** $$\frac{\frac{1}{\sin x} + \frac{1}{\cos x}}{\frac{1}{\sin x} - \frac{1}{\cos x}}$$ 3. **Find a common denominator for numerator and denominator:** - Numerator: $$\frac{\cos x + \sin x}{\sin x \cos x}$$ - Denominator: $$\frac{\cos x - \sin x}{\sin x \cos x}$$ 4. **Rewrite the entire expression:** $$\frac{\frac{\cos x + \sin x}{\sin x \cos x}}{\frac{\cos x - \sin x}{\sin x \cos x}} = \frac{\cos x + \sin x}{\cos x - \sin x}$$ 5. **Simplify the fraction:** The $$\sin x \cos x$$ terms cancel out because they appear in both numerator and denominator. 6. **Recognize the difference of squares in denominator:** Multiply numerator and denominator by the conjugate of the denominator to rationalize: $$\frac{\cos x + \sin x}{\cos x - \sin x} \times \frac{\cos x + \sin x}{\cos x + \sin x} = \frac{(\cos x + \sin x)^2}{\cos^2 x - \sin^2 x}$$ 7. **Expand numerator:** $$(\cos x + \sin x)^2 = \cos^2 x + 2 \sin x \cos x + \sin^2 x$$ 8. **Use Pythagorean identity:** $$\cos^2 x + \sin^2 x = 1$$ 9. **Substitute back:** $$\frac{1 + 2 \sin x \cos x}{\cos^2 x - \sin^2 x}$$ 10. **Recognize double angle formulas:** - $$\sin 2x = 2 \sin x \cos x$$ - $$\cos 2x = \cos^2 x - \sin^2 x$$ 11. **Rewrite expression using double angles:** $$\frac{1 + \sin 2x}{\cos 2x}$$ **Final answer:** $$\boxed{\frac{1 + \sin 2x}{\cos 2x}}$$