Giúp e giải mấy bài này Vs ạ e cần gấp lắm ạ

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
5,\\
a,\\
\frac{{1 - \cos \alpha }}{{{{\sin }^2}\alpha }} - \frac{1}{{1 + \cos \alpha }} = \frac{{1 - \cos \alpha }}{{1 - {{\cos }^2}\alpha }} - \frac{1}{{1 + \cos \alpha }}\\
 = \frac{{1 - \cos \alpha }}{{\left( {1 - \cos \alpha } \right)\left( {1 + \cos \alpha } \right)}} - \frac{1}{{1 + \cos \alpha }}\\
 = \frac{1}{{1 + \cos \alpha }} - \frac{1}{{1 + \cos \alpha }} = 0\\
b,\\
\frac{{1 - {{\sin }^2}\alpha .{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} - {\cos ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }} - {\sin ^2}\alpha  - {\cos ^2}\alpha \\
 = \frac{1}{{{{\cos }^2}\alpha }} - \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) = \frac{1}{{{{\cos }^2}\alpha }} - 1 = \frac{{1 - {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha \\
6,\\
a,\\
{\tan ^2}\alpha  - {\sin ^2}\alpha  = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} - {\sin ^2}\alpha  = {\sin ^2}\alpha \left( {\frac{1}{{{{\cos }^2}\alpha }} - 1} \right)\\
 = {\sin ^2}\alpha .\frac{{1 - {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\sin ^2}\alpha .\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\sin ^2}\alpha .{\tan ^2}\alpha \\
b,\\
\frac{{1 + {{\sin }^2}\alpha }}{{1 - {{\sin }^2}\alpha }} = \frac{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) + {{\sin }^2}\alpha }}{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) - {{\sin }^2}\alpha }}\\
 = \frac{{{{\cos }^2}\alpha  + 2{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = 1 + 2\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = 1 + 2{\tan ^2}\alpha \\
7,\\
\tan x.\cot x = \frac{{\sin x}}{{\cos x}}.\frac{{\cos x}}{{\sin x}} = 1\\
\frac{2}{{\tan \alpha  - 1}} + \frac{{\cot \alpha  + 1}}{{\cot \alpha  - 1}} = \frac{{2\left( {\cot \alpha  - 1} \right) + \left( {\cot \alpha  + 1} \right)\left( {\tan \alpha  - 1} \right)}}{{\left( {\tan \alpha  - 1} \right)\left( {\cot \alpha  - 1} \right)}}\\
 = \frac{{2\cot \alpha  - 2 + \cot \alpha .\tan \alpha  - \cot \alpha  + \tan \alpha  - 1}}{{\tan \alpha .\cot \alpha  - \tan \alpha  - \cot \alpha  + 1}}\\
 = \frac{{2\cot \alpha  - 2 + 1 - \cot \alpha  + \tan \alpha  - 1}}{{1 - \tan \alpha  - \cot \alpha  + 1}}\\
 = \frac{{\tan \alpha  + \cot \alpha  - 2}}{{2 - \tan \alpha  - \cot \alpha }} =  - 1
\end{array}\)