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Thread: Show that tgx+ctgx=2

  1. #1
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    Show that tgx+ctgx=2

    Given sinx+cosx=√2

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  2. #2
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    Re: Show that tgx+ctgx=2

    What is your question? What is tgx and what is ctgx? Are you trying to show that $\tan x + \cot x = 2$?

    Anyway, divide both sides by $\sqrt{2}$:

    $$\sin x \dfrac{1}{\sqrt{2}} + \cos x \dfrac{1}{\sqrt{2}} = 1$$

    $$\sin x \cos \dfrac{\pi}{4} + \cos x \sin \dfrac{\pi}{4} = 1$$

    $$\sin \left( x + \dfrac{\pi}{4} \right) = 1$$

    $$x+\dfrac{\pi}{4} = \dfrac{\pi}{2}+2n\pi$$

    $$x = \dfrac{\pi}{4}+2n\pi$$

    $$\tan x + \cot x = \tan \dfrac{\pi}{4} + \cot \dfrac{\pi}{4} = 1+1 = 2$$
    Last edited by SlipEternal; Apr 9th 2018 at 05:45 AM.
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  3. #3
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    Re: Show that tgx+ctgx=2

    Quote Originally Posted by SlipEternal View Post
    What is your question? What is tgx and what is ctgx? Are you trying to show that $\tan x + \cot x = 2$?
    Yes. I'm foreign and i forgot that they are written differently in english

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  4. #4
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    Re: Show that tgx+ctgx=2

    I updated my post.
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  5. #5
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    Re: Show that tgx+ctgx=2

    Quote Originally Posted by SlipEternal View Post
    What is your question? What is tgx and what is ctgx? Are you trying to show that $\tan x + \cot x = 2$?

    Anyway, divide both sides by $\sqrt{2}$:

    $$\sin x \dfrac{1}{\sqrt{2}} + \cos x \dfrac{1}{\sqrt{2}} = 1$$

    $$\sin x \cos \dfrac{\pi}{4} + \cos x \sin \dfrac{\pi}{4} = 1$$

    $$\sin \left( x + \dfrac{\pi}{4} \right) = 1$$

    $$x+\dfrac{\pi}{4} = \dfrac{\pi}{2}+2n\pi$$

    $$x = \dfrac{\pi}{4}+2n\pi$$

    $$\tan x + \cot x = \tan \dfrac{\pi}{4} + \cot \dfrac{\pi}{4} = 1+1 = 2$$
    Thanks!

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  6. #6
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    Re: Show that tgx+ctgx=2

    \begin{align*}
    \sin{(x)} + \cos{(x)} &= \sqrt2 \\
    \big(\sin{(x)} + \cos{(x)}\big)^2 &= 2 \\
    \sin^2{(x)} + 2\sin{(x)} \cos{(x)} + \cos^2{(x)} &= 2\big(\sin^2{(x)} + \cos^2{(x)}\big) \\
    2\sin{(x)} \cos{(x)} &= \sin^2{(x)} + \cos^2{(x)} \\
    2 &= \frac{\sin^2{(x)} + \cos^2{(x)}}{\sin{(x)} \cos{(x)}} \\
    &= \frac{\sin{(x)}}{\cos{(x)}} + \frac{\cos{(x)}}{\sin{(x)}} \\
    &= \tan{(x)} + \cot{(x)}
    \end{align*}
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