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

1. Rewrite the expression: \[ \frac{(\sin \theta+\cos \theta)^{2}}{\cos \theta} - \frac{1}{\cos \theta}. \] 2. Combine the terms over the common denominator: \[ \frac{(\sin \theta+\cos \theta)^{2} - 1}{\cos \theta}. \] 3. Expand the squared term in the numerator: \[ (\sin \theta+\cos \theta)^{2} = \sin^{2}\theta + 2\sin \theta\cos \theta + \cos^{2}\theta. \] 4. Utilize the Pythagorean identity \( \sin^{2}\theta + \cos^{2}\theta = 1 \): \[ \sin^{2}\theta + 2\sin \theta\cos \theta + \cos^{2}\theta = 1 + 2\sin \theta\cos \theta. \] 5. Substitute this back into the expression: \[ \frac{(1 + 2\sin \theta\cos \theta) - 1}{\cos \theta} = \frac{2\sin \theta\cos \theta}{\cos \theta}. \] 6. Cancel \( \cos \theta \) in the numerator and denominator (assuming \( \cos \theta \neq 0 \)): \[ 2\sin \theta. \] Thus, the given identity is proven: \[ \frac{(\sin \theta+\cos \theta)^{2}}{\cos \theta} - \frac{1}{\cos \theta} = 2\sin \theta. \]