Prove that:

$\mathrm{LHS}=\cot ^{2} x-\tan ^{2} x$

$=\frac{\cos ^{2} x}{\sin ^{2} x}-\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{\left(\cos ^{2} x\right)^{2}-\left(\sin ^{2} x\right)^{2}}{\sin ^{2} x \cos ^{2} x}$

$=\frac{\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x \cos ^{2} x}$

$=\frac{1 \times(\cos 2 x)}{\sin ^{2} x \cos ^{2} x} \quad\left(\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right)$

$=\frac{4 \cos 2 x}{4 \sin ^{2} x \cos ^{2} x}$

$=\frac{4(\cos 2 x)}{(\sin 2 x)^{2}}$

$=\frac{4(\cos 2 x)}{(\sin 2 x)} \times \frac{1}{(\sin 2 x)}$

$=4 \cot 2 x \operatorname{cosec} 2 x=\mathrm{RHS}$

Hence proved.