$\begin{array}{l}
4)\tan \left( {\dfrac{\pi }{4} + x} \right) - \tan \left( {\dfrac{\pi }{4} - x} \right)\\
= \dfrac{{\tan \dfrac{\pi }{4} + \tan x}}{{1 - \tan \dfrac{\pi }{4}\tan x}} - \dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}\\
= \dfrac{{1 + \tan x}}{{1 - \tan x}} - \dfrac{{1 - \tan x}}{{1 + \tan x}}\\
= \dfrac{{{{\left( {1 + \tan x} \right)}^2} - {{\left( {1 - \tan x} \right)}^2}}}{{1 - {{\tan }^2}x}}\\
= \dfrac{{{{\tan }^2}x + 1 + 2\tan x - {{\tan }^2}x + 2\tan x - 1}}{{1 - {{\tan }^2}x}}\\
= \dfrac{{4\tan x}}{{1 - {{\tan }^2}x}}\\
= 2\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right) = 2.\tan 2x\\
5)\dfrac{1}{{\sin 2x}} + \cot 2x = \dfrac{1}{{\sin 2x}} + \dfrac{{\cos 2x}}{{\sin 2x}}\\
= \dfrac{{1 + \cos 2x}}{{\sin 2x}} = \dfrac{{2{{\cos }^2}x - 1 + 1}}{{2\sin x\cos x}} = \dfrac{{2\cos x}}{{2\sin x}} = \dfrac{{\cos x}}{{\sin x}} = \cot x\\
6)\sin 3x = 3\sin x - 4{\sin ^3}x,\cos 3x = 4{\cos ^3}x - 3\cos x\\
\Leftrightarrow {\sin ^3}x = \dfrac{{3\sin x - \sin 3x}}{4},{\cos ^3}x = \dfrac{{\cos 3x + 3\cos x}}{4}\\
VT = \cos 3x{\sin ^3}x - \sin 3x{\cos ^3}x\\
= \cos 3x.\dfrac{{3\sin x - \sin 3x}}{4} - \dfrac{{\cos 3x + 3\cos x}}{4}\sin 3x\\
= \dfrac{1}{4}\left( {3\sin x\cos 3x - \sin 3x\cos 3x + \cos 3x\sin x + 3\cos x\sin 3x} \right)\\
= \dfrac{3}{4}\left( {\sin x\cos 3x + \sin 3x\cos x} \right) = \dfrac{3}{4}\sin \left( {3x + x} \right) = \dfrac{3}{4}\sin 4x
\end{array}$
