we have
SinA = 2/3
2 sec A/(1+ tan² A)
we know that
1+ tan² A = sec² A
substituting
2 sec A/(sec² A)--------- > 2 /sec A
remember that sec A=1/cos A
then
2 /sec A----------> 2 cos A
sin² A+cos² A=1-------> cos² A=1-sin² A-------> 1-(2/3)²
cos² A=1-(4/9)-----------> (9-4)/9-------------> cos² A=5/9 -------> cos A=(√5)/3
2 cos A=2*[(√5)/3)]=(2/3)√5
the answer is (2/3)√5
2 sec A/(1+ tan² A)=(2/3)√5