Cho biểu thức sau: tan x . tan(x + pi/3) + tan(x + pi/3).tan(x + 2pi/3) +
Đáp án: −3
Từ \(\tan \left( {a - b} \right) = \frac{{\tan a - \tan b}}{{1 + \tan a \cdot \tan b}} \Rightarrow \tan a \cdot \tan b = \frac{{\tan a - \tan b}}{{\tan \left( {a - b} \right)}} - 1\)
Áp dụng ta có:
\[\tan x \cdot \tan \left( {x + \frac{\pi }{3}} \right) = \frac{{\tan x - \tan \left( {x + \frac{\pi }{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1\]
\[\tan \left( {x + \frac{\pi }{3}} \right) \cdot \tan \left( {x + \frac{{2\pi }}{3}} \right) = \frac{{\tan \left( {x + \frac{\pi }{3}} \right) - \tan \left( {x + \frac{{2\pi }}{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1\]
\[\tan \left( {x + \frac{{2\pi }}{3}} \right) \cdot {\rm{tanx}} = \frac{{\tan \left( {x + \frac{{2\pi }}{3}} \right) - \tan x}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1\]
Khi đó \[\tan x \cdot \tan \left( {x + \frac{\pi }{3}} \right) + \tan \left( {x + \frac{\pi }{3}} \right) \cdot \tan \left( {x + \frac{{2\pi }}{3}} \right) + \tan \left( {x + \frac{{2\pi }}{3}} \right) \cdot {\rm{tanx}}\]
\[ = \frac{{\tan x - \tan \left( {x + \frac{\pi }{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1 + \frac{{\tan \left( {x + \frac{\pi }{3}} \right) - \tan \left( {x + \frac{{2\pi }}{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1 + \frac{{\tan \left( {x + \frac{{2\pi }}{3}} \right) - \tan x}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1 = - 3\]