Prove the identity: \( \frac{2 \tan x-\sin 2 x}{2 \sin ^{2} x}=\tan x \)

To prove the identity \( \frac{2 \tan x - \sin 2x}{2 \sin^2 x} = \tan x \), we start by rewriting the left-hand side. 1. Recall that \( \tan x = \frac{\sin x}{\cos x} \) and \( \sin 2x = 2 \sin x \cos x \). Thus, substituting these into the left-hand side gives: \[ \frac{2 \tan x - \sin 2x}{2 \sin^2 x} = \frac{2 \left(\frac{\sin x}{\cos x}\right) - 2 \sin x \cos x}{2 \sin^2 x} \] 2. Combine the terms in the numerator: \[ \frac{2 \sin x - 2 \sin x \cos^2 x}{2 \sin^2 x \cos x} = \frac{2 \sin x (1 - \cos^2 x)}{2 \sin^2 x} \] 3. Using the Pythagorean identity \( 1 - \cos^2 x = \sin^2 x \): \[ \frac{2 \sin x \sin^2 x}{2 \sin^2 x} = \sin x \] 4. Since \( \tan x = \frac{\sin x}{\cos x} \), simplifying gives us: \[ \frac{\sin x}{\sin^2 x} = \tan x, \] thus confirming our left-hand side simplifies to \( \tan x \), thereby proving the identity is valid. Hence, we conclude that: \[ \frac{2 \tan x - \sin 2x}{2 \sin^2 x} = \tan x. \] This completes the proof!