Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
E = \sin 36^\circ .\cos 6^\circ .\sin 126^\circ .\cos 84^\circ \\
= \sin 36^\circ .\cos 6^\circ .\cos \left( {90^\circ - 126^\circ } \right).\sin \left( {90^\circ - 84^\circ } \right)\\
= \sin 36^\circ .\cos 6^\circ .\cos \left( { - 36^\circ } \right).\sin 6^\circ \\
= \sin 36^\circ .\cos 6^\circ .\cos 36^\circ .\sin 6^\circ \\
= \frac{1}{4}.\left( {2\sin 36^\circ .\cos 36^\circ } \right).\left( {2\sin 6^\circ .cos6^\circ } \right)\\
= \frac{1}{4}.\sin 72^\circ .\sin 12^\circ \\
= \frac{1}{4}.\frac{{ - 1}}{2}.\left( {\cos \left( {72^\circ + 12^\circ } \right) - \cos \left( {72^\circ - 12^\circ } \right)} \right)\\
= - \frac{1}{8}.\left( {\cos 84^\circ - \cos 60^\circ } \right)\\
= \frac{{\frac{1}{2} - \cos 84^\circ }}{8} = \frac{{1 - 2\cos 84^\circ }}{{16}}\\
2,\\
\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{\cos \left( {90^\circ - x} \right)}}{{\sin \left( {90^\circ - x} \right)}} = \cot \left( {90^\circ - x} \right)\\
\tan x.\cot x = \frac{{\sin x}}{{\cos x}}.\frac{{\cos x}}{{\sin x}} = 1\\
\Rightarrow \tan x.\tan \left( {90^\circ - x} \right) = \tan x.\cot x = 1\\
A = \tan 1^\circ .\tan 2^\circ .tan3^\circ .tan4^\circ ......\tan 88^\circ .\tan 89^\circ \\
= \left( {\tan 1^\circ .\tan 89^\circ } \right).\left( {\tan 2^\circ .\tan 88^\circ } \right).......\left( {\tan 44^\circ .tan46^\circ } \right).\tan 45^\circ \\
= 1.1.1.1......1\\
= 1
\end{array}\)
