Given sinx+cosx=√2

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- Apr 9th 2018, 04:24 AMarchie13Show that tgx+ctgx=2
Given sinx+cosx=√2

Trimis de pe al meu SM-J530F folosind Tapatalk - Apr 9th 2018, 04:38 AMSlipEternalRe: 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 AMarchie13Re: Show that tgx+ctgx=2
- Apr 9th 2018, 04:46 AMSlipEternalRe: Show that tgx+ctgx=2
I updated my post.

- Apr 9th 2018, 04:48 AMarchie13Re: Show that tgx+ctgx=2
- Apr 9th 2018, 06:02 PMArchieRe: 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*}