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March 11th, 2008, 03:47 PM  #1 
Newbie Joined: Mar 2008 Posts: 1 Thanks: 0  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. Thanks, Rob 
March 11th, 2008, 07:12 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
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 04:42 PM. 
March 12th, 2008, 05:31 PM  #3 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
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. 
March 12th, 2008, 07:14 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
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 04:44 PM. 
March 12th, 2008, 08:22 PM  #5 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  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). 
March 13th, 2008, 01:07 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
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 04:46 PM. 
December 20th, 2016, 03:39 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,576 Thanks: 931 Math Focus: Elementary mathematics and beyond 
$\displaystyle \sec x+\tan x=1$ $\displaystyle \sec x+\tan x=\sec x\tan x$ $\displaystyle \tan x=\tan x$ 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 05:33 PM. 
December 20th, 2016, 04:35 PM  #8 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
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 $\displaystyle \begin{align*} x = 2\,k\,\pi \textrm{ where } k \in \mathbf{Z} \end{align*}$. 
December 20th, 2016, 08:43 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386  
December 21st, 2016, 05:13 PM  #10 
Member Joined: Dec 2016 From:  Posts: 54 Thanks: 10 
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{1z^{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 05:27 PM. 

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