E
Evaluate -32 + (2-6)(10).

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