Given $$ ab=1 $$ , $$ a≠ -1 $$ , $$ b≠ -1 $$ , find the value of $$ \dfrac{1}{1+a}+\dfrac{1}{1+b} $$

Question

Given $$ ab=1 $$ , $$ a≠ -1 $$ , $$ b≠ -1 $$ , find the value of $$ \dfrac{1}{1+a}+\dfrac{1}{1+b}  $$

Uniteasy Admin · Accepted Answer

$$ \dfrac{1}{1+a}+\dfrac{1}{1+b} $$
$$ =\dfrac{1+b+1+a}{(1+a)(1+b)} $$
$$ =\dfrac{2+a+b}{1+a+b+ab} $$
$$ =\dfrac{2+a+b}{1+a+b+1} $$
$$ =1 $$