I'm studying for a pre-calc test, and I have a practice problem thats completely stumped me, for any of the math geniuses out there, can you guys figure this one out? I have to verify that the following is a trigonometric identity:

[(cos x + 1)(tan x + sec x)]/(sec x + 1) = sin x + 1

To make it clearer to the human eye, I will write it like this as well -

(cos x + 1)(tan x + sec x) = sin x + 1
sec x + 1

Thanks so much!

I reject the notion that it takes a "math genius" to work these things out. (I know personally of, literally, hundreds of non-genius individuals who have mastered things up to and including freshman trigonometry, analytic geometry and, yes, even calculus.)

Math geniuses might not be able to tell you how to work it out (to them, it would be trivially obvious, I guess). I'm not sure that anyone can teach you how to prove that expressions are identical, but I think it's mostly a matter of familiarity and confidence.

I think it might take an understanding of fundamentals and, perhaps, a somewhat inquisitive nature. In other words: try to reduce expressions to more familiar forms and "noodle" around with them. If one approach doesn't lead to something that simplifies things, then try something else.

I would probably start by getting everything in terms of sin x and cos x.

As a notational convenience I will use S for sin x and C for cos x.

Now, noting that tan x = S/C and sec x = 1/C, I would work on the Left Hand Side. The idea is to fiddle around with it to see if we show it to be equal to S+1:

Regards,

Dave

"We can face our problem.
We can arrange such facts as we have
with order and method."

Hercule Poirot
--- in Murder on the Orient Express

I think that is kind of a bad problem, because the equality is not always true. The reason is that sec is undefined whenever x is an odd multiple of pi/2. The result will be that the entire left side is undefined while the right side is either 0 or 2 when you plug in any odd multiple of pi/2.

The reason I bring this up is that the next logical step from what Dave showed you is to cancel out the secants in the numerator or denominator. That will cause the undefined values of the left side to go away. Thus you actually change the value of the function at those specific points where it is undefined. Undefined divided by undefined is still undefined, not 1, which is what you get when you cancel out the secants.

Since the entire problem only works if you assume that x is not an odd multiple of pi/2, that is Ok. The places where you change the value of the function are not in the domain to begin with. However, canceling like that requires extreme caution.

I'm not bringing this up to confuse you (and I hope I didn't). I'm just trying to make you aware of a trap that a lot of people fall into when first learning to deal with trig identities. It's very easy to accidentally multiply or divide by zero, and doing so makes your whole proof wrong.

EDIT: To make my point more clear, consider the function y = (1 / x ) / ( 1 / x ). If x is not zero, you can just cancel out the 1/x on top and the 1/x on the bottom, which will always give you 1. However, if x = 0, the numerator and denominator are undefined. You can't cancel then. The function gives you undefined / undefined which is still undefined, not 1.

When you get to calculus you'll learn about limits, and then the rules will change, but until then, you just need to be careful.

EDIT 2: So this problem is like saying "show that (1/x)/(1/x) = 1." It's true accept at certain values of x. With your problem, those values are odd multiples of pi/2. In the example I showed you, it's zero.