# Show that tgx+ctgx=2

• Apr 9th 2018, 04:24 AM
archie13
Show that tgx+ctgx=2
Given sinx+cosx=√2

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• Apr 9th 2018, 04:38 AM
SlipEternal
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 $\sqrt{2}$:

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

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

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

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

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

$$\tan x + \cot x = \tan \dfrac{\pi}{4} + \cot \dfrac{\pi}{4} = 1+1 = 2$$
• Apr 9th 2018, 04:42 AM
archie13
Re: Show that tgx+ctgx=2
Quote:

Originally Posted by SlipEternal
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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• Apr 9th 2018, 04:46 AM
SlipEternal
Re: Show that tgx+ctgx=2
I updated my post.
• Apr 9th 2018, 04:48 AM
archie13
Re: Show that tgx+ctgx=2
Quote:

Originally Posted by SlipEternal
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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• Apr 9th 2018, 06:02 PM
Archie
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*}