cot² A - cot² B = $\dfrac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec² A - cosec² B

Solving,

$\Rightarrow \text{cot}^2 A - \text{cot}^2 B \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} - \dfrac{\text{cos}^2 B}{\text{sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 B - \text{cos}^2 B \text{ sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} \\[1em]$

By formula,

sin² θ = 1 - cos² θ

$\Rightarrow \dfrac{\text{cos}^2 A (1 - \text{ cos}^2 B) - \text{cos}^2 B (1 - \text{ cos}^2 A)}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2 A \text{cos}^2 B - \text{cos}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A - (1 - \text{sin}^2 B)}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{1 - 1 - \text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{-\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow -\dfrac{\text{sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} + \dfrac{\text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow -\dfrac{1}{\text{sin}^2 B} + \dfrac{1}{\text{sin}^2 A} \\[1em] \Rightarrow -\text{cosec}^2 B + \text{cosec}^2 A \\[1em] \Rightarrow \text{cosec}^2 A - \text{cosec}^2 B.$

Hence, proved that cot² A - cot² B = $\dfrac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec² A - cosec² B.