3.4 Prove the identity: \( \quad \frac{1}{(\cos \alpha+1)(\cos \alpha-1)}=\frac{-1}{\tan ^{2} \alpha \cdot \cos ^{2} \alpha} \)

This identity can be proved by manipulating both sides and using trigonometric identities. Starting with the left-hand side: \[ \frac{1}{(\cos \alpha + 1)(\cos \alpha - 1)} = \frac{1}{\cos^2 \alpha - 1} = \frac{1}{-(\sin^2 \alpha)} = -\frac{1}{\sin^2 \alpha} \] Next, we rewrite \(-\frac{1}{\sin^2 \alpha}\) in terms of \(\tan \alpha\) and \(\cos^2 \alpha\): \[ -\frac{1}{\sin^2 \alpha} = -\frac{\cos^2 \alpha}{\sin^2 \alpha \cdot \cos^2 \alpha} = -\frac{1}{\tan^2 \alpha \cdot \cos^2 \alpha} \] Thus, we have: \[ \frac{1}{(\cos \alpha + 1)(\cos \alpha - 1)} = \frac{-1}{\tan^2 \alpha \cdot \cos^2 \alpha} \] Both sides match, proving the identity.