2. Manipulate both LHS and RHS a. \( \frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{2 \tan x}{\cos x} \)

To solve this equation, let’s start by simplifying the left-hand side (LHS). We can find a common denominator for the two fractions: LHS: \[ \frac{1}{1-\sin x}-\frac{1}{1+\sin x} = \frac{(1+\sin x) - (1-\sin x)}{(1-\sin x)(1+\sin x)} = \frac{2\sin x}{1 - \sin^2 x} \] Using the Pythagorean identity \(1 - \sin^2 x = \cos^2 x\), we can rewrite the LHS: \[ LHS = \frac{2\sin x}{\cos^2 x} \] Now, let's simplify the right-hand side (RHS): \[ RHS = \frac{2 \tan x}{\cos x} = \frac{2 \sin x}{\cos^2 x} \] Both sides are equal, confirming that: \[ \frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{2 \tan x}{\cos x} \] is valid!