Proie the falowing identues 1- \( \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\sin ^{2} x \cdot \cos ^{2} x} \)

We start with the left‐hand side: \[ \frac{1}{(\cos x+1)(\cos x-1)} \] Notice that \[ (\cos x+1)(\cos x-1)=\cos^2 x-1 \] and using the Pythagorean identity \[ \sin^2 x+\cos^2 x=1 \quad\Longrightarrow\quad \cos^2 x-1=-\sin^2 x, \] we have \[ (\cos x+1)(\cos x-1)=-\sin^2 x. \] Thus, \[ \frac{1}{(\cos x+1)(\cos x-1)}=\frac{1}{-\sin^2 x}=-\frac{1}{\sin^2 x}. \] The given identity to prove is \[ \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\sin ^{2} x\cdot\cos ^{2} x}. \] Comparing our result \[ -\frac{1}{\sin^2 x} \] with the proposed right‐hand side \[ -\frac{1}{\sin^2 x\cdot\cos^2 x}, \] we see that they would be equal only if \[ \frac{1}{\sin^2 x}=\frac{1}{\sin^2 x\cdot\cos^2 x}, \] which requires \[ \cos^2 x=1. \] Since \(\cos^2 x=1\) only for \(x=k\pi\) (with \(k\) an integer), the equality \[ \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\sin ^{2} x\cdot\cos ^{2} x} \] holds only for those particular values of \(x\) and not in general. Thus, the identity as stated is not valid for all \(x\).