1. The problem is to verify or simplify the expression \( \frac{\sin x}{1-\sin x} + \frac{\sin x - 1}{1 + \sin x} = 4 \sec x \tan x \). 2. Start by simplifying the left-hand side (LHS). The LHS is a sum of two fractions: $$\frac{\sin x}{1-\sin x} + \frac{\sin x - 1}{1 + \sin x}$$ 3. Find a common denominator for the two fractions, which is \((1-\sin x)(1+\sin x) = 1 - \sin^2 x = \cos^2 x\) by the Pythagorean identity. 4. Rewrite each fraction with the common denominator: $$\frac{\sin x (1 + \sin x)}{\cos^2 x} + \frac{(\sin x - 1)(1 - \sin x)}{\cos^2 x}$$ 5. Expand the numerators: - First numerator: \(\sin x + \sin^2 x\) - Second numerator: \((\sin x - 1)(1 - \sin x) = \sin x - \sin^2 x - 1 + \sin x = 2 \sin x - \sin^2 x - 1\) 6. Add the numerators: $$\sin x + \sin^2 x + 2 \sin x - \sin^2 x - 1 = 3 \sin x - 1$$ 7. So the LHS becomes: $$\frac{3 \sin x - 1}{\cos^2 x}$$ 8. Now simplify the right-hand side (RHS): $$4 \sec x \tan x = 4 \cdot \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \frac{4 \sin x}{\cos^2 x}$$ 9. The equation to verify is: $$\frac{3 \sin x - 1}{\cos^2 x} = \frac{4 \sin x}{\cos^2 x}$$ 10. Multiply both sides by \(\cos^2 x\) (assuming \(\cos x \neq 0\)): $$3 \sin x - 1 = 4 \sin x$$ 11. Rearranged: $$3 \sin x - 1 - 4 \sin x = 0 \implies - \sin x - 1 = 0 \implies \sin x = -1$$ 12. Therefore, the original equation holds true only when \(\sin x = -1\). 13. The solution is that the equation is valid if and only if \(\sin x = -1\). Final answer: The given equation is true only at \(x = \frac{3\pi}{2} + 2k\pi\), where \(k\) is any integer.