\(\begin{array}{l}
G = \left( {{{\cot }^2}x + 2\cot x\tan x + {{\tan }^2}x} \right) - \left( {{{\cot }^2}x - 2\cot x\tan x + {{\tan }^2}x} \right)\\
= 4\cot x\tan x\\
= 4\\
\\
H = {\sin ^3}x\left( {1 + \dfrac{{\cos x}}{{\sin x}}} \right) + {\cos ^3}x\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)\\
= {\sin ^3}x\dfrac{{\sin x + \cos x}}{{\sin x}} + {\cos ^3}x\dfrac{{\cos x + \sin x}}{{\cos x}}\\
= {\sin ^2}x(\sin x + \cos x) + {\cos ^2}x(\sin x + \cos x)\\
= (\sin x + \cos x)({\sin ^2}x + {\cos ^2}x)\\
= \sin x + \cos x\\
\\
I = {\cot ^2}x(1 - {\sin ^2}x - 1) + 1\\
= {\cot ^2}x( - {\sin ^2}x) + 1\\
= \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}( - {\sin ^2}x) + 1\\
= - {\cos ^2}x + 1\\
= {\sin ^2}x\\
\\
J = \dfrac{{\cos 2x}}{{{{({{\sin }^2}x + {{\cos }^2}x)}^2} - 2{{\sin }^2}x{{\cos }^2}x - {{\sin }^2}x}} - 1\\
= \dfrac{{\cos 2x}}{{1 - {{\sin }^2}x - 2{{\sin }^2}x{{\cos }^2}x}} - 1\\
= \dfrac{{\cos 2x}}{{{{\cos }^2}x - 2{{\sin }^2}x{{\cos }^2}x}} - 1\\
= \dfrac{{\cos 2x}}{{{{\cos }^2}x(1 - 2{{\sin }^2}x)}} - 1\\
= \dfrac{{\cos 2x}}{{{{\cos }^2}x\cos 2x}} - 1\\
= \dfrac{1}{{{{\cos }^2}x}} - 1\\
= {\tan ^2}x
\end{array}\)
