Prove that \( \frac{1+2 \sin \theta \cdot \cos \theta}{\sin \theta+\cos \theta}=\sin \theta+\cos \theta \)

To prove the equation \[ \frac{1 + 2 \sin \theta \cdot \cos \theta}{\sin \theta + \cos \theta} = \sin \theta + \cos \theta, \] we can start by manipulating the left-hand side. 1. First, observe that \(2 \sin \theta \cos \theta\) can be rewritten using the double angle identity: \[ 2 \sin \theta \cos \theta = \sin(2\theta). \] 2. Now, let's express the numerator: \[ 1 + 2 \sin \theta \cos \theta = 1 + \sin(2\theta). \] So we have: \[ \frac{1 + \sin(2\theta)}{\sin \theta + \cos \theta}. \] 3. Now, letting \(x = \sin \theta + \cos \theta\), we can explore \(x^2\): \[ x^2 = (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = 1 + 2 \sin \theta \cos \theta. \] 4. Thus, \[ 1 + 2 \sin \theta \cos \theta = x^2, \] which can be substituted back into our equation: \[ \frac{x^2}{x}. \] 5. Simplifying gives us: \[ x, \] which is \(\sin \theta + \cos \theta\). 6. Therefore, we have shown that \[ \frac{1 + 2 \sin \theta \cos \theta}{\sin \theta + \cos \theta} = \sin \theta + \cos \theta. \] Hence, the original equation is proved. \[ \sin \theta + \cos \theta = \sin \theta + \cos \theta. \] This completes the proof.