Đáp án:

$\begin{array}{l}
1)P = \dfrac{{\sin a}}{{\cos a - 1}}\\
2)A = \cos a - \sin a\\
3)D = \cos a - \sin a
\end{array}$

Giải thích các bước giải:

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

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

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

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