In a triangle  $ABC,B=90°$, then the value of  $\tan \frac{A}{2}=$

$A+B+C=180$

$A+C=90(\because B=90°)$

$C=90-A$

$\tan \frac{A}{2}=\frac{b-c}{b+c}\cot \left(\frac{B-C}{2}\right)$

$=\frac{b-c}{b+c}\cot \left(\frac{90-90+A}{2}\right)$

$\tan \frac{A}{2}=\frac{b-c}{b+c}\cot \frac{A}{2}$

${\tan }^2\frac{A}{2}=\frac{b-c}{b+c}$

$\tan \frac{A}{2}=\sqrt{\frac{b-c}{b+c}}$