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March 11th, 2008, 04:47 PM   #1
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secx + tanx = 1 .... need help to simplify

Hello,
I was given the equation secx + tanx = 1 and I need to solve it (so I get an answer like pi/6 or something like that). I have tried several times to no avail and was hoping someone here could help. Any help would be greatly appreciated.

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Rob
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March 11th, 2008, 08:12 PM   #2
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The equation implies tan²x = (1 - sec x)² = 1 - 2sec x + sec²x = 2 - 2sec x + tan²x, so sec x = 1, and so cos x = 1, which is easy to solve.

Last edited by skipjack; December 20th, 2016 at 05:42 PM.
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March 12th, 2008, 06:31 PM   #3
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Here's my solution:
tan(x) + sec(x) = 1
sin(x)/cos(x) + 1/cos(x) = 1
(sin(x) + 1)/(cos(x)) = 1
cos(x) = sin(x) + 1
cos(x) = (+-)sqrt(1 - cos^2(x)) + 1 (Trig identity sin^2(x) + cos^2(x) = 1, hence sin(x) = (+-)sqrt(1 - cos^2(x)).)
cos(x) - 1 = (+-)sqrt(1 - cos^2(x))
cos^2(x) - 2cos(x) + 1 = 1 - cos^2(x)
0 = 2cos^2(x) - 2cos(x)
0 = cos(x)(cos(x) - 1)
cos(x) ?= 0 and/or cos(x) - 1 ?= 0
Since the original equation has cos(x) on denominator on fraction, we can say that cos(x) =/= 0, and cos(x) - 1 = 0, hence cos(x) = 1, and plug this in the original equation, and we get sin(x) + 1 = 1, so tan(x) = 0, hence we also get sec(x) = 1.
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March 12th, 2008, 08:14 PM   #4
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Dividing sec²x - tan²x = 1 by the given equation gives sec x - tan x = 1, and adding that to the original equation and then halving gives sec x = 1.

Last edited by skipjack; December 20th, 2016 at 05:44 PM.
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March 12th, 2008, 09:22 PM   #5
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Using trig identity:
1 + tan^2(x) = sec^2(x)

(sec(x) + tan(x))^2
= 1
= sec^2(x) + 2sec(x)tan(x) + tan^2(x)
= 1 + tan^2(x) + tan^2(x) + 2sec(x)tan(x)

0
= 2tan(x)(tan(x) + sec(x))

0
= tan(x)(tan(x) + sec(x))

sin(x)/cos(x) ?= tan(x) ?= 0 and/or tan(x) + sec(x) ?= 0
If sin(x) = 0 and cos(x) =/= 0, tan(x) = 0
sin(x)/cos(x) = -1/cos(x), where cos(x) =/= 0, then sin(x) = -1, but won't work because when sin(x) = -1, then cos(x) has to be 0, but cannot let be equal to zero, as following arithmetic rules on denominators of fractions.

Hence, we can say that sin(x) = 0 and cos(x) =/= 0, then tan(x) = 0. Plug this in the original equation, and we get
sec(x) + tan(x)
= 1
= sec(x).
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March 13th, 2008, 02:07 AM   #6
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1 = tan x + sec x = (sin x + 1)/cos x = (-cos(pi/2 + x) + 1)/sin(pi/2 + x) = 2sin²(pi/4 + x/2)/(2sin(pi/4 + x/2)cos(pi/4 + x/2)) = tan(pi/4 + x/2).
Hence pi/4 + x/2 = pi/4 + kpi (where k is an integer), so x = 2kpi.

Last edited by skipjack; December 20th, 2016 at 05:46 PM.
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December 20th, 2016, 04:39 PM   #7
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$\displaystyle \sec x+\tan x=1$

$\displaystyle \sec x+\tan x=\sec x-\tan x$

$\displaystyle \tan x=-\tan x$

which is only true when $x$ is an integer multiple of $2\pi$.



$\displaystyle \sec x+\tan x=1\quad[1]$

$\displaystyle \sec x+\tan x=(\sec x+\tan x)(\sec x-\tan x)\implies\sec x-\tan x=1\quad[2]$

$\displaystyle [1]+[2]\Rightarrow2\sec x=2\implies\sec x=1\implies x=2k\pi,k\in\mathbf{Z}$

(This method is used in skipjack's post #4)

Last edited by greg1313; December 20th, 2016 at 06:33 PM.
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December 20th, 2016, 05:35 PM   #8
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A couple of things to point out.

First of all, we should always think of the implied domains of the functions. As both $\displaystyle \begin{align*} \sec{(x)} \end{align*}$ and $\displaystyle \begin{align*} \tan{(x)} \end{align*}$ are defined through division of $\displaystyle \begin{align*} \cos{(x)} \end{align*}$, that means that the function won't be defined where $\displaystyle \begin{align*} \cos{(x)} = 0 \end{align*}$, i.e. where $\displaystyle \begin{align*} x = \frac{\left( 2\,n + 1 \right) \pi}{2} \end{align*}$, where $\displaystyle \begin{align*} n \in \mathbf{Z} \end{align*}$.

Second, at times it is unavoidable, but if you require squaring an equation in order to solve it, you can bring in extraneous solutions. Once you have gained a solution, you actually need to check if they all actually satisfy the original equation.

This is why I generally prefer to not square equations if I can avoid it.

$\displaystyle \begin{align*} \sec{(x)} + \tan{(x)} &= 1 \\ \frac{1}{\cos{(x)}} + \frac{\sin{(x)}}{\cos{(x)}} &= 1 \\ \frac{1 + \sin{(x)}}{\cos{(x)}} &= 1 \\ 1 + \sin{(x)} &= \cos{(x)} \\ \left[ 1 + \sin{(x)} \right] ^2 &= \cos^2{(x)} \\ 1 + 2\sin{(x)} + \sin^2{(x)} &= \cos^2{(x)} \\ 1 + 2\sin{(x)} + \sin^2{(x)} &= 1 - \sin^2{(x)} \\ 2\sin{(x)} + 2\sin^2{(x)} &= 0 \\ 2\sin{(x)}\left[ 1 + \sin{(x)} \right] &= 0 \end{align*}$

Case 1:

$\displaystyle \begin{align*} \sin{(x)} &= 0 \\ x &= m\,\pi \textrm{ where } m \in \mathbf{Z} \end{align*}$

Now as I said, as we have squared the equation, we need to check whether all of these solutions work in the original equation, so

$\displaystyle \begin{align*} \sec{ \left( m\,\pi \right) } + \tan{ \left( m\,\pi \right) } &= \frac{1}{\cos{ \left( m\,\pi \right) }} + 0 \\ &= \frac{1}{\left( -1 \right) ^m} \\ &= \left( -1 \right) ^m \end{align*}$

So only the times when m is even will give a solution to the original equation.

Thus a solution to the equation is $\displaystyle \begin{align*} 2\,k\,\pi \end{align*}$ where $\displaystyle \begin{align*} k \in \mathbf{Z} \end{align*}$.

Case 2:

$\displaystyle \begin{align*} 1 + \sin{(x)} &= 0 \\ \sin{(x)} &= -1 \\ x &= \frac{ \left( 4\,p - 1 \right) \pi}{2} \textrm{ where } p \in \mathbf{Z} \end{align*}$

But we have already established that any odd multiple of $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ is not in the domain of the original equation, so none of these solutions work.

So the complete solution to the equation $\displaystyle \begin{align*} \sec{(x)} + \tan{(x)} = 0 \end{align*}$ is $\displaystyle \begin{align*} x = 2\,k\,\pi \textrm{ where } k \in \mathbf{Z} \end{align*}$.
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December 20th, 2016, 09:43 PM   #9
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Quote:
Originally Posted by greg1313 View Post
. . . which is only true when $x$ is an integer multiple of $2\pi$.
$\tan x = -\tan x$ when $x$ is a multiple of $\pi$, but $\sec x = 1$ is also needed, which is true only when $x$ is an integer multiple of $2\pi$.
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December 21st, 2016, 06:13 PM   #10
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by inspection, you can inmediatly see that $x=0$ is a solution of such an equation. On the other hand
\begin{eqnarray}
\frac{1}{\cos x}+\frac{\sin x}{\cos x}=1 \to 1+\sin x = \cos x
\end{eqnarray}
Now call $z=\sin x$ and use that $1=\sin^{2} x + \cos^{2}x$. Then we have:
\begin{eqnarray}
1+z=\sqrt{1-z^{2}}\to z^{2}+z=0
\end{eqnarray}
The two solutions of that are:
\begin{eqnarray}
x_{1}&=&2 n\pi \\
x_{2}&=&(\frac{3\pi}{2} + 2\pi n)
\end{eqnarray}
where $n$ is any integer $n\geq 0$. EDIT: However, the solution $x_{2}$ doesnt satisfy the equation as can be easily proved. This has to do with the fact that, raising the power of $z$ to 2, we allow for an extra solution that in principle doesnt need to satisfy the equation.

Last edited by nietzsche; December 21st, 2016 at 06:27 PM.
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