Prove that \[ \cot^2x-\tan^2x=4\cot2x\cosec2x \]

Proof: Start with the left-hand side: \[ LHS=\cot^2x-\tan^2x \] Convert into sine and cosine form: \[ =\frac{\cos^2x}{\sin^2x}-\frac{\sin^2x}{\cos^2x} \] Taking LCM: \[ =\frac{\cos^4x-\sin^4x}{\sin^2x\cos^2x} \] Using \[ a^4-b^4=(a^2-b^2)(a^2+b^2) \] we get \[ =\frac{(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)}{\sin^2x\cos^2x} \] Using \[ \cos^2x+\sin^2x=1 \] and \[ \cos2x=\cos^2x-\sin^2x \] therefore, \[ LHS=\frac{\cos2x}{\sin^2x\cos^2x} \] Using \[ \sin2x=2\sin x\cos x \] Squaring both sides: \[ \sin^22x=4\sin^2x\cos^2x \] Thus, \[ \sin^2x\cos^2x=\frac{\sin^22x}{4} \] Substituting: \[ LHS = \frac{\cos2x}{\frac{\sin^22x}{4}} \] \[ = \frac{4\cos2x}{\sin^22x} \] \[ = 4\left(\frac{\cos2x}{\sin2x}\right)\left(\frac{1}{\sin2x}\right) \] \[ = 4\cot2x\cosec2x \] Hence proved, \[ \boxed{ \cot^2x-\tan^2x=4\cot2x\cosec2x } \]