Answer
See proof below.
Work Step by Step
Since, $sec (A+B)=\dfrac{1}{\cos(A+B)}$
Apply the sum formula : $\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{RHS}$ as follows:
$\text{LHS}=\dfrac{\csc a \csc B}{\cot a \cot B -1} \\=\dfrac{(\dfrac{1}{\sin a}) \times (\dfrac{1}{\sin B})}{(\dfrac{\cos a }{\sin a} ) \times (\dfrac{\cos B}{\sin B})-1}\\=\dfrac{1}{(\cos a) (\cos B)-(\sin a) (\sin B)}\\=\dfrac{1}{\cos (a+B)}\\=\sec (a+B) \\=\text{ LHS}$