\( \left|\begin{array}{l}\text { 3) Prove that } \\ \qquad \frac{\sin 2 \theta}{\sin \theta}=4 \cos \theta-\frac{\cos 2 \theta+1}{\cos \theta}\end{array}\right| \)

We start with the given identity: \[ \frac{\sin 2\theta}{\sin \theta} = 4 \cos \theta - \frac{\cos 2\theta + 1}{\cos \theta}. \] **Step 1. Express \(\sin 2\theta\) in terms of \(\sin \theta\) and \(\cos \theta\):** Recall the double-angle formula for sine: \[ \sin 2\theta = 2 \sin \theta \cos \theta. \] Thus, the left-hand side becomes: \[ \frac{\sin 2\theta}{\sin \theta} = \frac{2 \sin \theta \cos \theta}{\sin \theta} = 2 \cos \theta, \] assuming \(\sin \theta \neq 0\). **Step 2. Simplify the expression on the right-hand side:** Observe the term \(\cos 2\theta + 1\). Recall the double-angle formula for cosine: \[ \cos 2\theta = 2 \cos^2 \theta - 1. \] Thus, \[ \cos 2\theta + 1 = (2 \cos^2 \theta - 1) + 1 = 2 \cos^2 \theta. \] The right-hand side therefore becomes: \[ 4 \cos \theta - \frac{\cos 2\theta + 1}{\cos \theta} = 4 \cos \theta - \frac{2 \cos^2 \theta}{\cos \theta} = 4 \cos \theta - 2 \cos \theta = 2 \cos \theta. \] **Step 3. Compare both sides:** We have determined that: \[ \frac{\sin 2\theta}{\sin \theta} = 2 \cos \theta \] and \[ 4 \cos \theta - \frac{\cos 2\theta + 1}{\cos \theta} = 2 \cos \theta. \] Thus, both sides are equal: \[ 2 \cos \theta = 2 \cos \theta. \] This completes the proof.