# sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

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• Aug 21st 2012, 09:25 PM
Sjm2015
sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)
sec(t) = √2 cot(t) >0. Find all values of t in the interval [0,2).

I've been looking at examples of similar problems like this but I'm just not getting it. Would sec(t)=√2 be flipped as cos(t)=1/√2? Any help would be great. Lost as to what next step would be.
• Aug 21st 2012, 09:36 PM
Prove It
Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)
Quote:

Originally Posted by Sjm2015
sec(t) = √2 cot(t) >0. Find all values of t in the interval [0,2).

I've been looking at examples of similar problems like this but I'm just not getting it. Would sec(t)=√2 be flipped as cos(t)=1/√2? Any help would be great. Lost as to what next step would be.

\displaystyle \begin{align*} \sec{t} &= \sqrt{2}\cot{t} \\ \frac{1}{\cos{t}} &= \frac{\sqrt{2}\cos{t}}{\sin{t}} \\ \sin{t} &= \sqrt{2}\cos^2{t} \\ \sin{t} &= \sqrt{2}\left( 1 - \sin^2{t}\right) \\ \sin{t} &= \sqrt{2} - \sqrt{2}\sin^2{t} \\ \sqrt{2}\sin^2{t} + \sin{t} - \sqrt{2} &= 0 \end{align*}

Let \displaystyle \begin{align*} x = \sin{t} \end{align*} to give \displaystyle \begin{align*} \sqrt{2}\,x^2 + x - \sqrt{2} = 0 \end{align*}, solve for \displaystyle \begin{align*}x \end{align*}, then solve for \displaystyle \begin{align*} t \end{align*}.
• Aug 21st 2012, 09:58 PM
Sjm2015
Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)
Oh shoot, there was suppose to be a period seperating the two. sec(t) =√2. cot(t) >0. But thank you, that helped me with another problem lol.
• Aug 21st 2012, 10:23 PM
Prove It
Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)
Quote:

Originally Posted by Sjm2015
Oh shoot, there was suppose to be a period seperating the two. sec(t) =√2. cot(t) >0. But thank you, that helped me with another problem lol.

I don't see how that makes any difference to what I wrote. Is the RHS \displaystyle \begin{align*} \sqrt{2} \cdot \cot{t} \end{align*} or \displaystyle \begin{align*} \sqrt{2 \cdot \cot{t}} \end{align*}, or something else entirely?
• Aug 22nd 2012, 05:13 AM
skeeter
Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)
one equation, one conditional inequality ...

$\sec{t} = \sqrt{2}$

$\cot{t} > 0$
• Aug 22nd 2012, 06:45 AM
HallsofIvy
Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)
If it is, in fact, as skeeter suggests, to find t such that cot(t)> 0 and $sec(t)= \sqrt{2}$ Then $cos(t)= \frac{1}{sec(t)}= \frac{1}{\sqrt{2}}$ That is true for $t= \frac{\pi}{4}$ and $t= \frac{7\pi}{4}$ but only the first is less than 2 and has cot(t)> 0.