If a2+b2 = c2 , then 1logc+ab + 1logc-ab = ?

Given  $a2 + b2 = c2$

Now   $\frac{1}{\log c+ab } + \frac{1}{\log c-ab}$

=  $\log bc+a + \log bc-a$

=  $\log bc2-a2$

=  $2\log bb = 2$