Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
{\tan ^2}x - {\sin ^2}x = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - {\sin ^2}x = {\sin ^2}x\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right) = {\sin ^2}x.\frac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}} = {\sin ^2}x.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\sin ^2}x.{\tan ^2}x\\
b,\\
\frac{{\sin x + \cos x - 1}}{{\sin x - \cos x + 1}} = \frac{{\sin x + \left( {\cos x - 1} \right)}}{{\sin x - \left( {\cos x - 1} \right)}}\\
 = \frac{{{{\left[ {\sin x + \left( {\cos x - 1} \right)} \right]}^2}}}{{\left( {\sin x - \left( {\cos x - 1} \right)} \right)\left( {\sin x + \left( {\cos x - 1} \right)} \right)}}\\
 = \frac{{{{\sin }^2}x + 2.\sin x.\left( {\cos x - 1} \right) + {{\left( {\cos x - 1} \right)}^2}}}{{{{\sin }^2}x - {{\left( {\cos x - 1} \right)}^2}}}\\
 = \frac{{{{\sin }^2}x + 2\sin x\left( {\cos x - 1} \right) + {{\cos }^2}x - 2\cos x + 1}}{{{{\sin }^2}x - {{\cos }^2}x + 2\cos x - 1}}\\
 = \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x.\left( {\cos x - 1} \right) - 2\cos x + 1}}{{\left( {{{\sin }^2}x - 1} \right) - {{\cos }^2}x + 2\cos x}}\\
 = \frac{{1 + 2\sin x.\left( {\cos x - 1} \right) - 2\cos x + 1}}{{ - {{\cos }^2}x - {{\cos }^2}x + 2\cos x}}\\
 = \frac{{1 + \sin x\left( {\cos x - 1} \right) - \cos x}}{{ - {{\cos }^2}x + \cos x}}\\
 = \frac{{\left( {\cos x - 1} \right)\left( {\sin x - 1} \right)}}{{\cos x\left( {1 - \cos x} \right)}}\\
 = \frac{{1 - \sin x}}{{\cos x}} = \frac{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}{{\cos x\left( {1 + \sin x} \right)}}\\
 = \frac{{1 - {{\sin }^2}x}}{{\cos x.\left( {1 + \sin x} \right)}} = \frac{{{{\cos }^2}x}}{{\cos x\left( {1 + \sin x} \right)}} = \frac{{\cos x}}{{1 + \sin x}}\\
d,\\
\left( {1 + \tan x} \right).{\cos ^2}x + \left( {1 + \cot x} \right).{\sin ^2}x\\
 = \left( {1 + \frac{{\sin x}}{{\cos x}}} \right).{\cos ^2}x + \left( {1 + \frac{{\cos x}}{{\sin x}}} \right).{\sin ^2}x\\
 = \left( {\cos x + \sin x} \right).\cos x + \left( {\sin x + \cos x} \right).\sin x\\
 = {\cos ^2}x + \sin x.\cos x + {\sin ^2}x + \sin x.\cos x\\
 = {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x\\
 = {\left( {\sin x + \cos x} \right)^2}\\
e,\\
\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cot }^2}x - {{\tan }^2}x}} = \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{\frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = \frac{{{{\sin }^2}x.{{\cos }^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}{{{{\cos }^4}x - {{\sin }^4}x}}\\
 = \frac{{{{\sin }^2}x.{{\cos }^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}{{\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right)}}\\
 = {\sin ^2}x.{\cos ^2}x\,\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2} = 1} \right)
\end{array}\)