chứng minh đẳng thức: tg(a-b)+tg(b-c)+tg(c-a)=tg(a-b)tg(b-c)tg(c-a)

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chứng minh đẳng thức:
tg(a-b)+tg(b-c)+tg(c-a)=tg(a-b)tg(b-c)tg(c-a)

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Anna 12 phút 2021-09-10T12:56:01+00:00 2 Answers 0 views 0

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    0
    2021-09-10T12:57:26+00:00

    Ta có:

    $\quad \tan(x+y+z)=\tan[(x+y) + z]$

    $\to \tan(x+y+z)=\dfrac{\tan(x+y)+\tan z}{1 – \tan(x+y)\tan z}$

    $\to \tan(x+y+z)=\dfrac{\dfrac{\tan x + \tan y}{1 -\tan x.\tan y} + \tan z}{1 – \dfrac{\tan x +\tan y}{1-\tan x.\tan y}\cdot\tan z}$

    $\to \tan(x+y+z)=\dfrac{\tan x + \tan y +\tan z – \tan x.\tan y.\tan z}{1- \tan x.\tan y-\tan y.\tan z – \tan z.\tan x}$

    Đặt $\begin{cases}x = A – B\\y = B – C\\z = C – A\end{cases}$

    Ta được:

    $\quad\tan(x + y + z) = \tan(A-B + B – C + C -A)$

    $\to \tan(x+y+z)= \tan0$

    $\to \tan(x+y+z)= 0$

    $\to \dfrac{\tan x + \tan y +\tan z – \tan x.\tan y.\tan z}{1- \tan x.\tan y-\tan y.\tan z – \tan z.\tan x} = 0$

    $\to \tan x + \tan y +\tan z – \tan x.\tan y.\tan z = 0$

    $\to \tan x + \tan y +\tan z = \tan x.\tan y.\tan z$

    Hay $\tan(A-B) + \tan(B-C) + \tan(C-A) = \tan(A-B).\tan(B-C).\tan(C-A)$

    0
    2021-09-10T12:57:41+00:00

    Đáp án:

     

    Giải thích các bước giải: Hi, một cách để tham khảo thêm

    Công thức $: \dfrac{tanx + tany}{1 – tanxtany} = tan(x + y)$

    Với $ x = a – b; y = b – c  ⇒ x + y = a – c = – (c – a)$ thì:

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

    $ ⇔ tan(a – b) + tan(b – c) = – tan(c – a) + tan(a – b)tan(b – c)tan(c – a)$

    $ ⇔ tan(a – b) + tan(b – c) + tan(c – a) = tan(a – b)tan(b – c)tan(c – a) (đpcm)$

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