1 / (x/0) = 0...wait, isn't this undefined...?

Question

There was a recent question regarding a trigonometric equation, 1/sec(x) + 1/csc(x) = 1.  Other posters as well as myself disproved this equation.  I had an idea this equation wasn't true for any real x, but I was hesitant to say that was the case.  I looked into various math websites expecting to get a message saying 'no solutions found'.  Instead, I'm getting x = 0, π/2, and all of their coterminal equivalents as valid solutions.  But if that's the case, that would mean 1 / (1/0) = 0 as sec(x) is undefined at π/2 + nπ and csc(x) is undefined at nπ, if n is an integer.  How in the world is 1 / (1/0), or for that matter 1 / (x/0) equal to 0,  I thought these are undefined...??  I must be missing something...

冷眼旁觀 · Accepted Answer

(1/0) is undefined.

If (1/0) is the denominator of a fraction, the fraction is defined and equal to 0.
a/(1/0) = a × (0/1) = 0
where a is any real number.

However, if (1/0) is the numerator of a fraction, the fraction is undefined.
(1/0)/a = (1/0) × (1/a) = 1/0 = undefined
where a is any real number.

Jeffrey K · Answer

This is a technical mathematical point. The function y = 1/(1/x) is undefined when x is zero because division by zero is not allowed. But the limit as x approaches zero of 1/(1/x) is indeed zero. 
Another words, when x is almost zero, this function is almost zero. But x is never allowed to be exactly zero.

ted s · Answer

No , you missed nothing....1 / sec x + 1 / csc x = 1 demands that x ╫ nπ / 2.....the problem is cos x + sin x = 1 ??..square both sides to get 1 + sin 2x = 1 ===> sin 2x = 0 ===> x = nπ / 2....which is NOT in the domain.....thus no solution.....and nobody stated that ???

Captain Matticus, LandPiratesInc · Answer

So because you don't understand limits, you fail to grasp the concepts presented.  Good job!  You've revealed your ignorance!

? · Answer

1/sec(x) + 1/csc(x) = 1

=> cos x + sin x = 1

Divide throughout by cos x:

cos x / cos x + sin x / cos x = 1/ cos x
1 + tan x = sec x

Square both sides of the equation:
(1 + tan x)^2 = sec^2 x
1 + 2 tan x + tan^2 x = sec^2 x
2 tan x + 1 + tan^2 x = sec^2 x

But there is an identity: 1 + tan^2 x = sec^2 x

So 2 tan x + sec^2 x = sec^2 x

2 tan x = 0
tan x = 0
x = arctan (0) = 0, 2 pi (first and fourth quadrants)

billrussell42 · Answer

1/sec(x) + 1/csc(x) = 1
that is the same as 
cosx + sinx = 1
and there is no problem with that. as you can see by the graphical solution below.

1 / (x/0) = 0...wait, isn't this undefined...?

6 Answers