Answer
See proof below.
Work Step by Step
Since, $sec (A-B)=\dfrac{1}{\cos(A -B)}$
Apply the difference 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:
$\text{LHS}=\dfrac{\sec a \sec B}{\tan a \tan B +1} \\=\dfrac{(\dfrac{1}{\cos a}) \times (\dfrac{1}{\cos B})}{(\dfrac{\sin a }{\cos a} ) \times (\dfrac{\sin B}{\cos B})+1}\\=\dfrac{1}{(\cos a) (\cos B)+(\sin a) (\sin B)}\\=\dfrac{1}{\cos (a -B)}\\=\sec (a-B) \\=\text{ LHS}$
