Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
A = \left( {1 - {{\sin }^2}x} \right).co{t^2}x + 1 - {\cot ^2}x = {\cot ^2}x\left( {1 - {{\sin }^2}x - 1} \right) + 1\\
 = {\cot ^2}x.\left( { - {{\sin }^2}x} \right) + 1 = \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}.\left( { - {{\sin }^2}x} \right) + 1 = 1 - {\cos ^2}x = {\sin ^2}x\\
B = {\sin ^2}x + {\sin ^2}x.{\tan ^2}x = {\sin ^2}x\left( {1 + {{\tan }^2}x} \right) = {\sin ^2}x.\left( {1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)\\
 = {\sin ^2}x.\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}} = {\sin ^2}x.\frac{1}{{{{\cos }^2}x}} = {\tan ^2}x\\
C = \frac{{2{{\cos }^2}x - 1}}{{\sin x + \cos x}} = \frac{{{{\cos }^2}x + \left( {{{\cos }^2}x - 1} \right)}}{{\sin x + \cos x}} = \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x + \cos x}}\\
 = \frac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{\sin x + \cos x}} = \cos x - \sin x\\
D = {\left( {\tan x + \cot x} \right)^2} - {\left( {\tan x - \cot x} \right)^2}\\
 = {\tan ^2}x + 2.\tan x.\cot x + {\cot ^2}x - {\tan ^2}x + 2\tan x.\cot x - {\cot ^2}x\\
 = 4\tan x.\cot x = 4.\frac{{\sin x}}{{\cos x}}.\frac{{\cos x}}{{\sin x}} = 4\\
E = \frac{{{{\cos }^2}x + {{\cos }^2}x.{{\cot }^2}x}}{{{{\sin }^2}x + {{\sin }^2}x.{{\tan }^2}x}} = \frac{{{{\cos }^2}x\left( {1 + {{\cot }^2}x} \right)}}{{{{\sin }^2}x\left( {1 + {{\tan }^2}x} \right)}}\\
 = \frac{{{{\cos }^2}x\left( {1 + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)}}{{{{\sin }^2}x.\left( {1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}} = \frac{{{{\cos }^2}x.\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}}}}{{{{\sin }^2}x.\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}}}} = \frac{{{{\cos }^4}x}}{{{{\sin }^4}x}} = {\cot ^4}x\\
F = \frac{{{{\sin }^2}x - {{\cos }^2}x + {{\cos }^4}x}}{{{{\cos }^2}x - {{\sin }^2}x + {{\sin }^4}x}} = \frac{{{{\sin }^2}x - {{\cos }^2}x\left( {1 - {{\cos }^2}x} \right)}}{{{{\cos }^2}x - {{\sin }^2}x\left( {1 - {{\sin }^2}x} \right)}}\\
 = \frac{{{{\sin }^2}x - {{\cos }^2}x.{{\sin }^2}x}}{{{{\cos }^2}x - {{\sin }^2}x.{{\cos }^2}x}} = \frac{{{{\sin }^2}x\left( {1 - {{\cos }^2}x} \right)}}{{{{\cos }^2}x\left( {1 - {{\sin }^2}x} \right)}} = \frac{{{{\sin }^2}x.{{\sin }^2}x}}{{{{\cos }^2}x.{{\cos }^2}x}} = {\tan ^4}x\\
G = \sqrt {\frac{{1 + \cos x}}{{1 - \cos x}}}  - \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}}  = \frac{{{{\sqrt {1 + \cos x} }^2} - {{\sqrt {1 - \cos x} }^2}}}{{\sqrt {1 - \cos x} .\sqrt {1 + \cos x} }}\\
 = \frac{{\left( {1 + \cos x} \right) - \left( {1 - \cos x} \right)}}{{\sqrt {1 - {{\cos }^2}x} }} = \frac{{2\cos x}}{{\sqrt {{{\sin }^2}x} }} = 2\cot x\\
H = \cos x - \sqrt {{{\tan }^2}x - {{\sin }^2}x} \\
 = \cos x - \sqrt {{{\sin }^2}x\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} \\
 = \cos x - \sqrt {{{\sin }^2}x.\frac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}} \\
 = \cos x - \sqrt {{{\sin }^2}x.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \\
 = \cos x + \frac{{{{\sin }^2}x}}{{\cos x}}\,\,\,\,\,\,\,\,\left( {x \in \left( {\frac{\pi }{2};\frac{{3\pi }}{2}} \right) \Rightarrow \cos x < 0} \right)\\
 = \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{\cos x}} = \frac{1}{{\cos x}}
\end{array}\)