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

To prove that \[ \frac{1 + 2 \sin \theta \cdot \cos \theta}{\sin \theta + \cos \theta} = \sin \theta + \cos \theta, \] we start by multiplying both sides of the equation by \(\sin \theta + \cos \theta\) to eliminate the denominator: \[ 1 + 2 \sin \theta \cdot \cos \theta = (\sin \theta + \cos \theta)(\sin \theta + \cos \theta). \] The right side can be expanded using the formula for the square of a binomial: \[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta. \] Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can rewrite the equation: \[ (\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta. \] Thus, our equation simplifies to: \[ 1 + 2 \sin \theta \cos \theta = 1 + 2 \sin \theta \cos \theta. \] Since both sides are equal, we have proven the original identity: \[ \frac{1 + 2 \sin \theta \cdot \cos \theta}{\sin \theta + \cos \theta} = \sin \theta + \cos \theta. \] This concludes the proof.