Answer
$tan(\frac{A}{2})=\frac{3-\sqrt{5}}{2}$
Work Step by Step
Note$\;\;\;\;\;\;\;sinA=\frac{2}{3}\;\;\;;\;\;\;\;0^o\leq A\leq 90^o$
$sin^2(A)+cos^2(A)=1\;\;\;\;\;\;\;\;\;\;\;\;$ and $sin(A)=\frac{2}{3}$
$(\frac{2}{3})^2+cos^2(A)=1$
$cos^2(A)=\frac{5}{9}$
$cos(A)=\pm \frac{\sqrt{5}}{3}$
We know $0^o\leq A\leq 90^o$
$cos(A)=\frac{\sqrt{5}}{3}$
$tan(\frac{A}{2})= \frac{1+cos(A)}{sin(A)}$
$tan(\frac{A}{2})=\frac{1+\frac{\sqrt{5}}{3}}{\frac{2}{3}}=\frac{3-\sqrt{5}}{2}$
$tan(\frac{A}{2})=\frac{3-\sqrt{5}}{2}$
