Zero equals one? By using inverse trig identities, it can be shown that $sec(x)*cos(x)=1$. However, when $x=π/2$, the resulting function would be $$(1/0)*(0/1)=1,$$ and cross multiplying yields $$1*0=1.$$ I'm not sure where I went wrong in this example but there doesn't seem to be any apparent gaps in the graph of $sec(x)*cos(x)$. What am I doing wrong?
 A: *

*The domain of $\sec x$ does not include $\frac{\pi}{2}$; $\sec\frac{\pi}{2}$ is not defined. More information here.

*Even if you didn't know that, or forgot it, $\frac{1}{0}$ is not defined.

*Even if you forgot those things, 
$0\cdot\frac{1}{0} = \frac{0}{0}$
by your logic, which is undefined. 
A: The terms $1/0,0/0,\infty$ are indeterminate or undefined and cant be used for algebra eg $6\times 0=5\times 0$ cancelling zero we get $6=5$ so you get why they cant be used !
A: $m$ divided by $n$ is the specific number, $q$, such that $qn = m$.  So $m$ divided by $0$ is the specific number, $q$, such that $q*0 = m$.  There is no such number so division by 0 is impossible and 1/0 is undefined.
Now, trigonometry. You have a unit circle.  You have an angle $\theta$.  The points on the unit circle corresponding to the angle $\theta$ are $(x, y) = (\cos \theta, \sin \theta)$ and there is an inscribed right triangle corresponding to that angle with sides $x = \cos \theta$, $y = \sin \theta$ and hypotenuse = 1.  That's basic.
So... Draw a tangent line from the (x,y) point that is tangent to the circle.  Extend that line until it crosses the x-axis (!!Warning!! !!Warning!! Who ever told you it would cross the x-axis?) (Oh, shut up, voice in my head; most lines will cross the x-axis.)... extend that line until it crosses the x-axis.  This forms another right angle that is similar to the first.  The tangent line from $(x,y)$ to the x-axis coresponds to the $x = \cos \theta$ side of the first triangle.  The radius of the unit circle (from the origin to $(x,y)$) on the new triangle corresponds with the  $y = \sin \theta$ side of the original triangle. The x-axis from the origin to where the line intersects the x-axis corresponds to the hypotenuse of the original triangle.
Doing a bit of geometry with realize the scaling factor of this new triangle is $1/\cos \theta$.  So the distance of the tangent point to the x-axis is $\sin \theta /\cos \theta$.  We call that $\tan \theta$ (because it's the length of a tangent chord!!!).  And the length from the $(0,0)$ origin to where the tangent line intersects the x-axis is $1/\cos \theta$.  We call that $\sec \theta$ (because it is the length of a secant chord !... wait, what the heck is a "secant chord"?  .. ).
Let me repeat that as I'm sure your eyes are glazing over.
$\sec \theta$ is the distance from the origin to where the tangent line intersects the x-axis.
Okay, so let $\theta = \pi/2$.  $\sec \pi/2$ is the distance from the origin to where the tangent line intersects the x-axis.  But the tangent line is parallel to the x-axis!!!!  The tangent line never intersects the x-axis and there is no such distance!
(Maybe I should have listened to that voice in my head.  Wouldn't be the first time.  Of course, there was the time when it told me to... well, never mind.)
So $\sec \pi/2$ is undefined.
Just like $1/0$ is undefined.  
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There are many similarities.  $\sec \theta = 1/\cos \theta$ so $\sec \pi/2$ would $= 1/0$ if such a thing were possible.  
Also notice as $\theta$ gets closer and closer to $\pi/2$ the tangent line gets larger and larger and $\sec \theta$ gets larger and larger and approaches infinity.
Likewise if you are dividing $m$ by $n$ and $n$ gets smaller and smaller and closer to 0 then $m/n$ gets larger and approaches infinity.
