Đáp án:

$\begin{array}{l}
A = 1 - 2{\sin ^2}x\\
 = \left( {1 - \sqrt 2 \sin x} \right)\left( {1 + \sqrt 2 \sin x} \right)\\
B = 3 - {\tan ^2}x\\
 = \left( {\sqrt 3  + \tan x} \right)\left( {\sqrt 3  - \tan x} \right)\\
C = \sin x.\cos x + \cos x + \sin x + 1\\
 = \cos x\left( {\sin x + 1} \right) + \sin x + 1\\
 = \left( {\sin x + 1} \right)\left( {\cos x + 1} \right)\\
D = \sin x.\cos x + {\cos ^2}x - 1\\
 = \sin x.\cos x + {\cos ^2}x - {\sin ^2}x - {\cos ^2}x\\
 = \sin x\left( {\cos x - \sin x} \right)\\
E = {\sin ^2}x + \sin x.\cos x - 1\\
 = \sin x.\cos x - {\cos ^2}x\\
 = \cos x\left( {\sin x - \cos x} \right)\\
F = 1 + \tan x + \sin x + \cos x\\
 = 1 + \frac{{\sin x}}{{\cos x}} + \sin x + \cos x\\
 = \left( {\sin x + \cos x} \right).\left( {\frac{1}{{\cos x}} + 1} \right)\\
G = \tan x - \cot x + \sin x + \cos x\\
 = \frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\sin x}} + \sin x + \cos x\\
 = \frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\cos x.sinx}} + \sin x + \cos x\\
 = \left( {\cos x + \sin x} \right)\left( {\frac{{\sin x - \cos x}}{{\cos x.\sin x}} + 1} \right)\\
H = \cos x.{\tan ^2}x - 1 - \cos x\\
 = \cos x.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1 - \cos x\\
 = \frac{{{{\sin }^2}x}}{{\cos x}} - \left( {\cos x + 1} \right)\\
 = \frac{{1 - {{\cos }^2}x}}{{\cos x}} - \left( {\cos x + 1} \right)\\
 = \left( {\cos x + 1} \right).\left( {\frac{{1 - \cos x}}{{\cos x}} - 1} \right)
\end{array}$