Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
4,\\
\dfrac{{1 - 2{{\cos }^2}x}}{{{{\sin }^2}x.{{\cos }^2}x}} = \dfrac{{\left( {1 - {{\cos }^2}x} \right) - {{\cos }^2}x}}{{{{\sin }^2}x.{{\cos }^2}x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x.{{\cos }^2}x}}\\
 = \dfrac{1}{{{{\cos }^2}x}} - \dfrac{1}{{{{\sin }^2}x}} = \left( {\dfrac{1}{{{{\cos }^2}x}} - 1} \right) - \left( {\dfrac{1}{{{{\sin }^2}x}} - 1} \right)\\
 = \dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}} - \dfrac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}} = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = {\tan ^2}x - {\cot ^2}x\\
2,\\
\dfrac{{\cos x}}{{1 + \sin x}} + \tan x = \dfrac{{\cos x}}{{1 + \sin x}} + \dfrac{{\sin x}}{{\cos x}}\\
 = \dfrac{{{{\cos }^2}x + \sin x.\left( {1 + \sin x} \right)}}{{\left( {1 + \sin x} \right).\cos x}}\\
 = \dfrac{{\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + \sin x}}{{\left( {1 + \sin x} \right).\cos x}}\\
 = \dfrac{{1 + \sin x}}{{\left( {1 + \sin x} \right).\cos x}} = \dfrac{1}{{\cos x}}\\
6,\\
\dfrac{{{{\tan }^2}x - {{\tan }^2}y}}{{{{\tan }^2}x.{{\tan }^2}y}} = \dfrac{1}{{{{\tan }^2}y}} - \dfrac{1}{{{{\tan }^2}x}} = \dfrac{1}{{\dfrac{{{{\sin }^2}y}}{{{{\cos }^2}y}}}} - \dfrac{1}{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}\\
 = \dfrac{{{{\cos }^2}y}}{{{{\sin }^2}y}} - \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = \left( {\dfrac{{{{\cos }^2}y}}{{{{\sin }^2}y}} + 1} \right) - \left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} + 1} \right)\\
 = \dfrac{{{{\cos }^2}y + {{\sin }^2}y}}{{{{\sin }^2}y}} - \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\sin }^2}x}}\\
 = \dfrac{1}{{{{\sin }^2}y}} - \dfrac{1}{{{{\sin }^2}x}} = \dfrac{{{{\sin }^2}x - {{\sin }^2}y}}{{{{\sin }^2}x.{{\sin }^2}y}}
\end{array}\)