Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
11,\\
\dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{{{\left( {1 + {{\tan }^2}x} \right)}^2}}}\\
= \dfrac{{4.\dfrac{{\sin x}}{{\cos x}}\left( {1 - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}{{{{\left( {1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}^2}}}\\
= \dfrac{{4.\dfrac{{\sin x}}{{\cos x}}.\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}}}{{{{\left( {\dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}^2}}}\\
= \dfrac{{\dfrac{{4\sin x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}{{{{\cos }^3}x}}}}{{{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}^2}}}\\
= 4\sin x.\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\\
= 2.\left( {2\sin x.\cos x} \right).\cos 2x\\
= 2.\sin 2x.\cos 2x\\
= \sin 4x\\
12,\\
\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 - \tan x.\tan y}}\\
\Rightarrow \tan x - \tan y = \left( {1 - \tan x.\tan y} \right).\tan \left( {x - y} \right)\\
\tan 3\alpha - tan2\alpha - tan\alpha - tan\alpha .tan2\alpha .tan3\alpha \\
= \left( {\tan 3\alpha - tan\alpha } \right) - \tan 2\alpha - \tan \alpha .\tan 2\alpha .\tan 3\alpha \\
= \left( {1 - \tan 3\alpha .\tan \alpha } \right).\tan 2\alpha - \tan 2\alpha - \tan \alpha .\tan 2\alpha .\tan 3\alpha \\
= \tan 2\alpha - \tan 3\alpha .\tan \alpha .\tan 2\alpha - \tan 2\alpha - \tan \alpha .\tan 2\alpha .\tan 3\alpha \\
= 0
\end{array}\)
