A confusing trigonometric question

[Hells_Kitchen]

A confusing trigonometric question!

Hi guys, I had a question. I was given the following equality. There was no problem statement that came with it (i.e. it did not say solve the equation or prove the identity, or anything of this nature)

sec x + csc x / tan x + cot x = sin x + cos x

The person that handed it to me said that he solved it whatever that means. Now, if this was an identity (i.e. true for all permissible values of x) than it should be true for say x=45 degrees. Clearly it is not!

Then I considered it as an equation (i.e. true for SOME permissible values of x). I studied the function f(x) = LHS - RHS and I saw that this function does not attain the value 0. So even as an equation, there is no solution.

Am I doing something wrong here, or is the other person doing something wrong?
Thanks guys!

 

[HallsofIvy]

Please, please, please use paretheses to make yourself clear! I assume you actually mean (sec x + csc x )/ (tan x + cot x) = sin x + cos x. What you wrote would, using the standard rules of algebra, be sec x + (csc x / tan x)+ cot x = sin x + cos x
but that is clearly not true.

My general rule, not always the simplest, is to start by replacing all functions by sine and cosine. Here, you have, on the left,
[tex]\frac{\frac{1}{cos x}+ \frac{1}{sin x}}{\frac{sin x}{cos x}+ \frac{cos x}{sin x}}[/tex].

Now multiply both numerator and denominator by of the main fraction by sin x cos x.

 

[Hells_Kitchen]

hmmm. Someone who is taking algebra gave me this equation and the way he had written it was exactly this:
sec x + (csc x / tan x)+ cot x = sin x + cos x

He could not solve it for some reason, but this equation cannot be solved as it is written.

On the other hand, (sec x + csc x )/ (tan x + cot x) = sin x + cos x is an identity which is easily proved as you suggested above. I was arguing with him and I was telling him that he must have copied the problem wrong. So I see that he forgot parantheses which turned an easy problem into a nightmare.

ok then, sorry about the hassle. Thanks for the quick response. But I did mean the equation without parantheses because that equation does not have a solution if you try to solve it!