Prove the identity $\;$ $\dfrac{b^2 - 3b - \left(b - 1\right) \sqrt{b^2 - 4} + 2}{b^2 + 3b - \left(b + 1\right) \sqrt{b^2 - 4} + 2} \sqrt{\dfrac{b + 2}{b - 2}} = \dfrac{1 - b}{1 + b}$
LHS $= \dfrac{b^2 - 3b - \left(b - 1\right) \sqrt{b^2 - 4} + 2}{b^2 + 3b - \left(b + 1\right) \sqrt{b^2 - 4} + 2} \sqrt{\dfrac{b + 2}{b - 2}}$
$= \dfrac{b^2 - b - 2b - \left(b - 1\right) \sqrt{b^2 - 4} + 2}{b^2 + b + 2b - \left(b + 1\right) \sqrt{b^2 - 4} + 2} \times \dfrac{\sqrt{b + 2}}{\sqrt{b - 2}}$
$= \dfrac{b \left(b - 1\right) - 2 \left(b - 1\right) - \left(b - 1\right) \sqrt{b^2 - 4}}{b \left(b + 1\right) + 2 \left(b + 1\right) - \left(b + 1\right) \sqrt{b^2 - 4}} \times \dfrac{\sqrt{b + 2}}{\sqrt{b - 2}}$
$= \dfrac{\left(b - 1\right) \left[b - 2 - \sqrt{b^2 - 4}\right]}{\left(b + 1\right) \left[b + 2 - \sqrt{b^2 - 4}\right]} \times \dfrac{\sqrt{b + 2}}{\sqrt{b - 2}}$
$= \dfrac{\left(b - 1\right) \left[\sqrt{b - 2} \times \sqrt{b - 2} - \sqrt{b + 2} \times \sqrt{b - 2}\right]}{\left(b + 1\right) \left[\sqrt{b + 2} \times \sqrt{b + 2} - \sqrt{b + 2} \times \sqrt{b - 2}\right]} \times \dfrac{\sqrt{b + 2}}{\sqrt{b - 2}}$
$= \dfrac{\left(b - 1\right) \left(\sqrt{b - 2}\right) \left(\sqrt{b - 2} - \sqrt{b + 2}\right)}{\left(b + 1\right) \left(\sqrt{b + 2}\right) \left(\sqrt{b + 2} - \sqrt{b - 2}\right)} \times \dfrac{\sqrt{b + 2}}{\sqrt{b - 2}}$
$= \dfrac{-\left(b - 1\right)}{b + 1}$
$= \dfrac{1 - b}{1 + b}$
$= RHS$