If (tanx+secx)2=A+sinxBsinx,\displaystyle {\left({\tan{{x}}}+{\sec{{x}}}\right)}^{{2}}={\frac{{{A}+{\sin{{x}}}}}{{{B}-{\sin{{x}}}}}}, then
A=\displaystyle {A}=   ,
B=\displaystyle {B}=   .