Question 532241
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ +\ 1\ =\ \sec(x)]


Use the definitions of tangent and secant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin(x)}{\cos(x)}\ +\ 1\ =\ \frac{1}{\cos(x)}]


Add the fractions in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin(x)\ +\ \cos(x)}{\cos(x)}\ =\ \frac{1}{\cos(x)}]


Equate the numerators for the equal denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ +\ \cos(x)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ 1\ -\ \cos(x)]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(x)\ =\ 1\ -\ 2\cos(x)\ +\ \cos^2(x)]


Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2(x)\ =\ 1\ -\ 2\cos(x)\ +\ \cos^2(x)]


Collect terms in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2(x)\ -\ 2\cos(x)\ =\ 0]


Factor


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos(x)\left(\cos(x)\ -\ 1\right)\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ 1]


But going back to the original equation, *[tex \Large \cos(x)] is in the denominator, so discard the *[tex \Large \cos(x)\ =\ 0] root.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ 1\ \Rightarrow\ x\ =\ \cos^{-1}(1)\ =\ 0]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>