If θ = cot^ - 1 α, φ = cot^ - 1 β and αβ = - 1 (α > β), then the...

$∵ \cot^{-1}\alpha = \theta ⇒ \alpha = cot \theta........... \left(i\right)$
Also, $\cot^{-1}\beta = \phi⇒ \beta =cot \phi........... \left(ii\right)$
Now, given $\alpha\beta = -1$

So, from ( i ) & ( ii ), we get,
$cot \theta.cot \phi = - 1 ⇒ tan \theta tan \phi = - 1......... \left(iii\right)$
$\tan \left(\theta - \phi\right) = \dfrac{tan \theta - tan \phi}{1 + tan \theta tan \phi}............\left(iv\right)$
$\theta -ϕ = \pm \dfrac{\pi}{2} (from ( iii ) and ( iv ) )$
But, $\alpha> \beta⇒ \theta< \phi ⇒ \theta - \phi = - \dfrac{\pi}{2}$