Đáp án:

$A = \left\{ \begin{array}{l}
\sqrt 2 .\tan a,0 < a < \dfrac{\pi }{2}\\
\dfrac{{ - \sqrt 2 }}{{\cos a}},\dfrac{\pi }{2} < a < \pi 
\end{array} \right.$

Giải thích các bước giải:

ĐKXĐ: $a \ne \dfrac{\pi }{2}$

 Ta có:

$\begin{array}{l}
A = \dfrac{{\sin \left( {\dfrac{\pi }{4} + \dfrac{a}{2}} \right)}}{{\sqrt {1 - \sin a} }} - \dfrac{{\sin \left( {\dfrac{\pi }{4} - \dfrac{a}{2}} \right)}}{{\sqrt {1 + \sin a} }}\\
 = \dfrac{{\dfrac{1}{{\sqrt 2 }}\cos \dfrac{a}{2} + \dfrac{1}{{\sqrt 2 }}\sin \dfrac{a}{2}}}{{\sqrt {1 - \sin a} }} - \dfrac{{\dfrac{1}{{\sqrt 2 }}\cos \dfrac{a}{2} - \dfrac{1}{{\sqrt 2 }}\sin \dfrac{a}{2}}}{{\sqrt {1 + \sin a} }}\\
 = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}{{\sqrt {1 - \sin a} }} - \dfrac{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}}{{\sqrt {1 + \sin a} }}} \right)\\
 = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}{{\sqrt {{{\sin }^2}\left( {\dfrac{a}{2}} \right) - 2\sin \dfrac{a}{2}\cos \dfrac{a}{2} + {{\cos }^2}\left( {\dfrac{a}{2}} \right)} }} - \dfrac{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}}{{\sqrt {{{\sin }^2}\left( {\dfrac{a}{2}} \right) + 2\sin \dfrac{a}{2}\cos \dfrac{a}{2} + {{\cos }^2}\left( {\dfrac{a}{2}} \right)} }}} \right)\\
 = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}{{\sqrt {{{\left( {\sin \dfrac{a}{2} - \cos \dfrac{a}{2}} \right)}^2}} }} - \dfrac{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}}{{\sqrt {{{\left( {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right)}^2}} }}} \right)\\
 = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}{{\left| {\sin \dfrac{a}{2} - \cos \dfrac{a}{2}} \right|}} - \dfrac{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}}{{\left| {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right|}}} \right)
\end{array}$

Lại có:

$0 < a < \pi  \Rightarrow 0 < \dfrac{a}{2} < \dfrac{\pi }{2} \Rightarrow \cos \dfrac{a}{2} > 0;\sin \dfrac{a}{2} > 0$

$ + )TH1:0 < \dfrac{a}{2} < \dfrac{\pi }{4} \Rightarrow \cos \dfrac{a}{2} > \sin \dfrac{a}{2} > 0$

Khi đó:

$\begin{array}{l}
A = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}} - \dfrac{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}}{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}} \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\dfrac{{{{\left( {\cos \dfrac{a}{2} + \sin \dfrac{a}{2}} \right)}^2} - {{\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)}^2}}}{{\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)\left( {\cos \dfrac{a}{2} + \sin \dfrac{a}{2}} \right)}}\\
 = \dfrac{1}{{\sqrt 2 }}.\dfrac{{4\cos \dfrac{a}{2}\sin \dfrac{a}{2}}}{{{{\cos }^2}\left( {\dfrac{a}{2}} \right) - {{\sin }^2}\left( {\dfrac{a}{2}} \right)}}\\
 = \dfrac{1}{{\sqrt 2 }}.\dfrac{{2\sin a}}{{\cos a}}\\
 = \sqrt 2 .\tan a
\end{array}$

$ + )TH2:\dfrac{\pi }{4} < \dfrac{a}{2} < \dfrac{\pi }{2} \Rightarrow \sin \dfrac{a}{2} > \cos \dfrac{a}{2} > 0$

Khi đó:

$\begin{array}{l}
A = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}{{\sin \dfrac{a}{2} - \cos \dfrac{a}{2}}} - \dfrac{{\cos \dfrac{a}{2} - \sin \dfrac{a}{2}}}{{\cos \dfrac{a}{2} + \sin \dfrac{a}{2}}}} \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\dfrac{{{{\left( {\cos \dfrac{a}{2} + \sin \dfrac{a}{2}} \right)}^2} + {{\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)}^2}}}{{\left( {\sin \dfrac{a}{2} - \cos \dfrac{a}{2}} \right)\left( {\cos \dfrac{a}{2} + \sin \dfrac{a}{2}} \right)}}\\
 = \dfrac{1}{{\sqrt 2 }}.\dfrac{{2\left( {{{\cos }^2}\left( {\dfrac{a}{2}} \right) + {{\sin }^2}\left( {\dfrac{a}{2}} \right)} \right)}}{{{{\sin }^2}\left( {\dfrac{a}{2}} \right) - {{\cos }^2}\left( {\dfrac{a}{2}} \right)}}\\
 = \dfrac{1}{{\sqrt 2 }}.\dfrac{2}{{ - \cos a}}\\
 = \dfrac{{ - \sqrt 2 }}{{\cos a}}
\end{array}$

Vậy $A = \left\{ \begin{array}{l}
\sqrt 2 .\tan a,0 < a < \dfrac{\pi }{2}\\
\dfrac{{ - \sqrt 2 }}{{\cos a}},\dfrac{\pi }{2} < a < \pi 
\end{array} \right.$

$ A = \dfrac{{\sin \left( {\dfrac{\pi }{4} + \dfrac{a}{2}} \right)}}{{\sqrt {1 - \sin a} }} - \dfrac{{\sin \left( {\dfrac{\pi }{4} - \dfrac{a}{2}} \right)}}{{\sqrt {1 + \sin a} }}\\ A = \dfrac{{\sin \left( {\dfrac{\pi }{4} + \dfrac{a}{2}} \right)\sqrt {1 + \sin a}  - \sin \left( {\dfrac{\pi }{4} - \dfrac{a}{2}} \right)\sqrt {1 - \sin a} }}{{\sqrt {1 - {{\sin }^2}a} }}\\ A = \dfrac{{\dfrac{{\sqrt 2 }}{2}\left( {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right)\sqrt {1 + \sin a}  - \dfrac{{\sqrt 2 }}{2}\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)\sqrt {1 - \sin a} }}{{\cos a}}\\ A = \dfrac{{\dfrac{{\sqrt 2 }}{2}\left( {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right)\sqrt {{{\left( {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right)}^2}}  - \dfrac{{\sqrt 2 }}{2}\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)\sqrt {{{\left( {\sin \dfrac{a}{2} - \cos \dfrac{a}{2}} \right)}^2}} }}{{\cos a}}\\ \left[ \begin{array}{l} A = \dfrac{{\dfrac{{\sqrt 2 }}{2}{{\left( {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right)}^2} - \dfrac{{\sqrt 2 }}{2}{{\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)}^2}}}{{\cos a}}\\ A = \dfrac{{\dfrac{{\sqrt 2 }}{2}{{\left( {\sin \dfrac{a}{2} + \cos \dfrac{a}{2}} \right)}^2} + \dfrac{{\sqrt 2 }}{2}{{\left( {\cos \dfrac{a}{2} - \sin \dfrac{a}{2}} \right)}^2}}}{{\cos a}} = \dfrac{{\sqrt 2 }}{{\cos a}} \end{array} \right.\\ \left[ \begin{array}{l} A = \dfrac{{2\sqrt 2 \sin \dfrac{a}{2}\cos \dfrac{a}{2}}}{{\cos a}} = \dfrac{{\sqrt 2 \sin a}}{{\cos a}} = \sqrt 2 \tan a\left( {0 < a < \dfrac{\pi }{2}} \right)\\ A = \dfrac{{\sqrt 2 }}{{\cos a}}\left( {\dfrac{\pi }{2} < a < \pi} \right) \end{array} \right. $