7.3 Consider the identity: \( \left\lvert\, \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x\right. \) 7.3.1 Prove the identity. 7.3.2 Determine the values of \( x \) for which this identity is undefined.

To prove the identity \( \left\lvert\, \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x\right. \), we'll start with the left side. Recall that \( \sin 2x = 2\sin x \cos x \) and \( \cos 2x = 2\cos^2 x - 1 \). We can rewrite: \[ \sin x + \sin 2x = \sin x + 2 \sin x \cos x = \sin x (1 + 2 \cos x) \] For the denominator, \( 1 + \cos x + \cos 2x = 1 + \cos x + (2\cos^2 x - 1) = \cos x + 2\cos^2 x \). Thus, we have the left-hand side as: \[ \frac{\sin x (1 + 2 \cos x)}{\cos x(1 + 2 \cos x)} = \frac{\sin x}{\cos x} = \tan x \] Now concerning the values where this identity is undefined, we consider the denominator \( 1 + \cos x + \cos 2x \). It is undefined when: \[ 1 + \cos x + (2\cos^2 x - 1) = 0 \implies \cos x + 2 \cos^2 x = 0 \] Factoring gives us \( \cos x (2 \cos x + 1) = 0 \), which results in: 1. \( \cos x = 0 \), corresponding to \( x = \frac{\pi}{2} + k\pi \) for any integer \( k \). 2. \( 2 \cos x + 1 = 0 \implies \cos x = -\frac{1}{2} \), leading to \( x = \frac{2\pi}{3} + 2k\pi \) or \( x = \frac{4\pi}{3} + 2k\pi \) for any integer \( k \). Therefore, the identity is undefined for: - \( x = \frac{\pi}{2} + k\pi \) - \( x = \frac{2\pi}{3} + 2k\pi \) - \( x = \frac{4\pi}{3} + 2k\pi \)