Help Needed: Problem from 2001 Lial Text - Can You Provide A Clue?

[Hsopitalist]

Homework Statement

((sec x - tan x)^2 +1)/(sec x cscx) - (tanx secx ) = 2 tan x

Relevant Equations

multiple trig identities

This is my first attempt to ask for help on here. I'm not in school (52 years old) but just exploring. This is a problem from an old Lial text from 2001.

I have worked on this problem for almost an hour now and just need some resolution. I have four pages of notes here. I have tried all kinds of substitutions and can't get a way to connect the two sides. It's number 48 on page 200 for those who have it. Can someone give me a clue for how to solve this?
Thanks.

 

[Hsopitalist]

Never mind, I got it!

 

[Mark44]

Homework Statement:: ((sec x - tan x)^2 +1)/(sec x cscx) - (tanx secx ) = 2 tan x

@Hsopitalist, just to be clear, is the identity you're trying to prove this (which is what you wrote):
$$\frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x} - \tan x\sec x = 2\tan x$$
or is it this one?
$$\frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x - \tan x\sec x} = 2\tan x$$
Edit: Both equations are edited to correctly reflect the original equation.
We've had a lot of members who wrote something like x^2 - 1/x - 1, when what they meant was (x^2 - 1)/(x - 1).
An expression written as x^2 - 1/x - 1 would usually be interpreted to mean $x^2 - \frac 1 x - 1$.

 

[Hsopitalist]

Mark44

Interesting, thanks for that observation. I guess I should've put another set of parenthesis around the entire denominator. Which brings me to how were you able to write what you wrote? I like that way better than the method I used.

Sean

 

[Mark44]

I gather it was this one:
$$\frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x - \tan x\sec x} = 2\tan x$$

Here's the TeX script I wrote, $$\frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x - \tan x\sec x} = 2\tan x$$

We have a LaTeX guide -- a link to it appears in the lower left corner.

 

[Hsopitalist]

$$\frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x - \tan x\sec x} = 2\tan x$$

Oh how cool is that. Thanks!

 

[Hsopitalist]

While I'm at it, I know there are programs like MATLAB and Mathematica. Is it worthwhile to start learning one of those or just save it until it comes up in a pedagogical format?

 

[archaic]

Never mind, I got it!

how so?

 

[Hsopitalist]

Archaic

Thanks for your input, this is why I love this website! My comment "how cool is that" was in reference to the fact that I had never used LaTeX before, not the equation itself.

The correct equation was in my original post, in the denominator there should have been "- tanx secx" whereas as I originally wrote it it seemed to be a term onto itself and not under the denominator.

 

[Mark44]

Both are wrong; just checked with desmos. You can also take $x=\frac{\pi}{4}$.

I miscopied two of the trig functions in the posted identity. I've gone back an edited my post to correct the equations.

 

[SammyS]

As it turns out, this expression is not an identity.

$\displaystyle \frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x - \tan x\sec x} = 2\tan x$

Modify it by replacing the last sec(x) with csc(x). The result is a trig identity.

$\displaystyle \frac{(\sec x - \tan x)^2 + 1}{\sec x\csc x - \tan x\csc x} = 2\tan x$