Rút gọn A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx

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Rút gọn
A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx

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  2. Đáp án:

     $A=-3$

    Giải thích các bước giải:

    Áp dụng công thức:

    $ \tan (a – b) = \dfrac{\tan a – \tan b}{1 + \tan a.\tan b}$

    $ ⇒ \tan (a – b).(1 + \tan a.\tan b) = \tan a – \tan b$ (*)

    $A=\tan x.\tan \left({x + \dfrac{π}{3}}\right) + \tan \left({x + \dfrac{π}{3}}\right).\tan \left({x + \dfrac{2π}{3}}\right) + \tan \left({x + \dfrac{2π}{3}}\right).\tan x$

    Áp dụng (*) ta có:

    +)

    $ \tan \dfrac{π}{3}.\left[{1 + \tan x.\tan \left({x + \dfrac{π}{3}}\right)}\right]$

    $ = \tan \left[{\left({x + \dfrac{π}{3}}\right) – x)}\right].\left[{1 + \tan x.\tan \left({x + \dfrac{π}{3}}\right)}\right]$

    $= \tan \left({x + \dfrac{π}{3}}\right) – \tan x$ (1)

    +)

    $\tan \dfrac{π}{3}.\left[{1 + \tan \left({x + \dfrac{π}{3}}\right).\tan \left({x + \dfrac{2π}{3}}\right)}\right]$

    $= \tan \left[{\left({x + \dfrac{2π}{3}}\right) – \left({x + \dfrac{π}{3}}\right)}\right].\left[{1 + \tan \left({x + \dfrac{π}{3}}\right).\tan \left({x + \dfrac{2π}{3}}\right)}\right] $

    $ = \tan \left({x + \dfrac{2π}{3}}\right) – \tan \left({x + \dfrac{π}{3}}\right)$ (2)

    +)

    $ \tan \dfrac{π}{3}.\left[{1 + \tan \left({x + \dfrac{2π}{3}}\right).\tan x}\right] $

    $ = \tan \dfrac{- 2π}{3}.\left[{1 + \tan \left({x + \dfrac{2π}{3}}\right).\tan x}\right] $

    $ = \tan \left[{x – \left({x + \dfrac{2π}{3}}\right)}\right].\left[{1 + \tan \left({x + \dfrac{π}{3}}\right).\tan x}\right]$

    $ = \tan x – \tan \left({x + \dfrac{2π}{3}}\right)$ (3)

    Lấy vế cộng vế phương trình (1) + (2) + (3) ta có:

    $ \tan \dfrac{π}{3}.(3+A) =0$

    $ \Leftrightarrow A = – 3$

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