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[tex]\displaystyle \\ \frac{\sin x}{1-\cos x} + \frac{\sin x}{1+\cos x} = \\ \\ =\frac{\sin x(1+\cos x)}{(1-\cos x)(1+\cos x)} + \frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)} = \\ \\ =\frac{\sin x+\sin x\cos x}{1-\cos^2x} + \frac{\sin x-\sin x\cos x}{1-\cos^2x} = \\ \\ =\frac{\sin x+\sin x\cos x + \sin x-\sin x\cos x}{1-\cos^2x} = \\ \\ =\frac{2\sin x }{\sin^2x} = \frac{2 }{\sin x} =2 \times \frac{1 }{\sin x} = 2 \times \csc x = \boxed{2\csc x }[/tex]



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