1. **Problem statement:** Prove the identity $$\frac{\tan x - \sin x}{2 \tan x} = \sin^2 \left(\frac{x}{2}\right)$$. 2. **Recall formulas:** - $$\tan x = \frac{\sin x}{\cos x}$$ - Double angle and half angle formulas: $$\sin^2 \left(\frac{x}{2}\right) = \frac{1 - \cos x}{2}$$ 3. **Rewrite the left side (LHS):** $$\frac{\tan x - \sin x}{2 \tan x} = \frac{\frac{\sin x}{\cos x} - \sin x}{2 \cdot \frac{\sin x}{\cos x}} = \frac{\frac{\sin x - \sin x \cos x}{\cos x}}{\frac{2 \sin x}{\cos x}}$$ 4. **Simplify numerator and denominator:** $$= \frac{\sin x (1 - \cos x)/\cos x}{2 \sin x / \cos x} = \frac{\sin x (1 - \cos x)}{\cos x} \times \frac{\cos x}{2 \sin x} = \frac{1 - \cos x}{2}$$ 5. **Recognize the right side (RHS):** $$\sin^2 \left(\frac{x}{2}\right) = \frac{1 - \cos x}{2}$$ 6. **Conclusion:** Since both sides simplify to $$\frac{1 - \cos x}{2}$$, the identity is proven. **Final answer:** $$\frac{\tan x - \sin x}{2 \tan x} = \sin^2 \left(\frac{x}{2}\right)$$.