Giải thích các bước giải:

$\begin{array}{l}
B2:\\
a)A = \dfrac{{1 - {{\sin }^2}\alpha }}{{1 - {{\cos }^2}\alpha }} + \tan \alpha \cot \alpha \\
 = \dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} + 1\\
 = {\cot ^2}\alpha  + 1\\
b)B = \dfrac{{1 + \cos \alpha }}{{\cos \alpha }}\left[ {1 - \dfrac{{{{\left( {1 - \cos \alpha } \right)}^2}}}{{{{\sin }^2}\alpha }}} \right]\\
 = \dfrac{{1 + \cos \alpha }}{{\cos \alpha }}\left[ {1 - \dfrac{{{{\left( {1 - \cos \alpha } \right)}^2}}}{{1 - {{\cos }^2}\alpha }}} \right]\\
 = \dfrac{{1 + \cos \alpha }}{{\cos \alpha }}\left[ {1 - \dfrac{{{{\left( {1 - \cos \alpha } \right)}^2}}}{{\left( {1 - \cos \alpha } \right)\left( {1 + \cos \alpha } \right)}}} \right]\\
 = \dfrac{{1 + \cos \alpha }}{{\cos \alpha }}\left[ {1 - \dfrac{{1 - \cos \alpha }}{{1 + \cos \alpha }}} \right]\\
 = \dfrac{{1 + \cos \alpha }}{{\cos \alpha }}.\dfrac{{2\cos \alpha }}{{1 + \cos \alpha }}\\
 = 2\\
c)C = \dfrac{{\tan \alpha  + \tan \beta }}{{\cot \alpha  + \cot \beta }}\\
 = \dfrac{{\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\sin \beta }}{{\cos \beta }}}}{{\dfrac{{\cos \alpha }}{{\sin \alpha }} + \dfrac{{\cos \beta }}{{\sin \beta }}}}\\
 = \dfrac{{\dfrac{{\sin \alpha \cos \beta  + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\dfrac{{\cos \alpha \sin \beta  + \sin \alpha \cos \beta }}{{\sin \alpha \sin \beta }}}}\\
 = \dfrac{{\sin \alpha \cos \beta  + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}.\dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \sin \beta  + \sin \alpha \cos \beta }}\\
 = \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}\\
 = \tan \alpha \tan \beta \\
B3:\\
D = {\sin ^8}\alpha  + {\cos ^8}\alpha  - 2{\left( {1 - {{\sin }^2}\alpha {{\cos }^2}\alpha } \right)^2}\\
 = {\left( {{{\sin }^4}\alpha  + {{\cos }^4}\alpha } \right)^2} - 2{\sin ^4}\alpha {\cos ^4}\alpha  - 2\left( {{{\sin }^4}\alpha {{\cos }^4}\alpha  - 2{{\sin }^2}\alpha {{\cos }^2}\alpha  + 1} \right)\\
 = {\left( {{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)}^2} - 2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right)^2} - 4{\sin ^4}\alpha {\cos ^4}\alpha  + 4{\sin ^2}\alpha {\cos ^2}\alpha  - 2\\
 = {\left( {1 - 2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right)^2} - 4{\sin ^4}\alpha {\cos ^4}\alpha  + 4{\sin ^2}\alpha {\cos ^2}\alpha  - 2\\
 = 4{\sin ^4}\alpha {\cos ^4}\alpha  - 4{\sin ^2}\alpha {\cos ^2}\alpha  + 1 - 4{\sin ^4}\alpha {\cos ^4}\alpha  + 4{\sin ^2}\alpha {\cos ^2}\alpha  - 2\\
 =  - 1
\end{array}$