Proving trigonomeric identities \( \frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{2 \tan x}{\cos x} \)

To prove the trigonometric identity \[ \frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{2 \tan x}{\cos x}, \] we will start by simplifying the left-hand side (LHS) of the equation. ### Step 1: Simplify the Left-Hand Side The LHS is \[ \frac{1}{1-\sin x} - \frac{1}{1+\sin x}. \] To combine these fractions, we need a common denominator, which is \((1 - \sin x)(1 + \sin x)\). Thus, we can rewrite the LHS as: \[ \frac{(1 + \sin x) - (1 - \sin x)}{(1 - \sin x)(1 + \sin x)}. \] ### Step 2: Simplify the Numerator Now, simplify the numerator: \[ (1 + \sin x) - (1 - \sin x) = 1 + \sin x - 1 + \sin x = 2\sin x. \] So, the LHS becomes: \[ \frac{2\sin x}{(1 - \sin x)(1 + \sin x)}. \] ### Step 3: Simplify the Denominator The denominator can be simplified using the difference of squares: \[ (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x. \] ### Step 4: Rewrite the Left-Hand Side Now we can rewrite the LHS as: \[ \frac{2\sin x}{\cos^2 x}. \] ### Step 5: Rewrite in Terms of Tangent Recall that \(\tan x = \frac{\sin x}{\cos x}\). Therefore, we can express the LHS as: \[ \frac{2\sin x}{\cos^2 x} = 2 \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = 2 \tan x \cdot \frac{1}{\cos x}. \] ### Step 6: Final Form Thus, we have: \[ \frac{2\sin x}{\cos^2 x} = \frac{2 \tan x}{\cos x}. \] ### Conclusion Since we have shown that the left-hand side simplifies to the right-hand side, we conclude that: \[ \frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{2 \tan x}{\cos x}. \] This proves the identity.