0. \( \frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=\frac{4 \tan x}{\cos x} \)

To solve the equation \[ \frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=\frac{4 \tan x}{\cos x}, \] we start by simplifying the left side. The common denominator for the two fractions on the left side is \((1-\sin x)(1+\sin x)\). Therefore, we can write the left side as: \[ \frac{(1+\sin x)^2 - (1-\sin x)^2}{(1-\sin x)(1+\sin x)}. \] We expand the numerator: \[ (1+\sin x)^2 - (1-\sin x)^2 = (1 + 2\sin x + \sin^2 x) - (1 - 2\sin x + \sin^2 x) = 2\sin x + 2\sin x = 4\sin x. \] Now, substituting this into the equation gives us: \[ \frac{4\sin x}{(1-\sin x)(1+\sin x)}. \] Next, recall that \( (1-\sin x)(1+\sin x) = 1 - \sin^2 x = \cos^2 x\). Hence, the left side can be rewritten as: \[ \frac{4\sin x}{\cos^2 x}. \] So now our equation is: \[ \frac{4\sin x}{\cos^2 x} = \frac{4\tan x}{\cos x}. \] We can simplify the right side: \[ \frac{4\tan x}{\cos x} = \frac{4\sin x / \cos x}{\cos x} = \frac{4\sin x}{\cos^2 x}. \] Since both sides of the equation are equal, we can see that the original equation is an identity for values of \(x\) for which both sides are defined and valid. Thus, the equality is satisfied for all \(x\) not equal to those values that make the denominators zero.