# Simplifying a trigonometry expression

1. Aug 25, 2013

### mileena

1. The problem statement, all variables and given/known data

Simplify:

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

2. Relevant equations

none

3. The attempt at a solution

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - tan x (1 -1) sec x ] /
sec2 x

[(sec x)(sec2 x) - 0 ] /
sec2 x

sec x

Thanks! I hope the above is correct, but it probably isn't.

2. Aug 25, 2013

### haruspex

That step is wrong. You've dropped a tan x. Try it again.
But I would start by getting rid of the denominator. What is another way of expressing 1/sec(θ)?

3. Aug 25, 2013

### HallsofIvy

Staff Emeritus
So
$$\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}$$?

Is there any particular reason for writing "$sec(x)sec^2(x)$" rather than $sec^3(x)$?

Where did that "(1- 1)" come from?
sec2 x

[(sec x)(sec2 x) - 0 ] /
sec2 x

sec x

Thanks! I hope the above is correct, but it probably isn't.[/QUOTE]
It's easy to check that it is NOT correct. If x= 30 degrees ($\pi/6$ radians), the original expression is $(2/3)\sqrt{3}+ 2/3$ while sec(x) is $(2/3)\sqrt{3}$.

4. Aug 25, 2013

### HallsofIvy

Staff Emeritus
So
$$\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}$$?

Is there any particular reason for writing "$sec(x)sec^2(x)$" rather than $sec^3(x)$?

Where did that "(1- 1)" come from?

It's easy to check that it is NOT correct. If x= 30 degrees ($\pi/6$ radians), the original expression is $(2/3)\sqrt{3}+ 2/3$ while sec(x) is $(2/3)\sqrt{3}$.

5. Aug 26, 2013

### mileena

Thanks haruspex, and sorry for my late reply. I have a hard time getting online in the evenings.

cos x = 1/sec x

So:

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
[(sec x)(sec x)]

(cos x)(cos x) [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]

cos2 x [sec3 x - (sec2 x tan x - sec x tan x)]

[sec3 x/sec2 x] - [[(sec x)(tan2 x - tan x)]/sec2 x]

sec x - [[(tan2 x - tan x)]/sec x]

sec x - [cos x[(tan2 x - tan x)]]

That's it. I don't know how to reduce this more.

Last edited: Aug 26, 2013
6. Aug 26, 2013

### janhaa

maybe you can simplify it this way:

$1-\left(\frac{\tan^2(x)-\tan(x)}{1+\tan^2(x)}\right)$

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

7. Aug 26, 2013

### mileena

That is the way the problem was presented to me on the homework!

I tried to distribute the tan x out of both factors!

8. Aug 26, 2013

### haruspex

There's a mistake in how you multiplied out the (tan x - 1)(sec x tan x) term.
Why flip back to dividing by sec x again? You can use sec x * cos x = 1 without having to turn it back into a division.

9. Aug 26, 2013

### mileena

Ok, I decided to redo the problem. Part of my problem is it's hard to check the work on here because I can't write fractions they way they are supposed to be written without complex computer operations.

Thank you harupex for the tip: sec x * cos x = 1

I knew they were inverses, but it didn't dawn on me to use that equation!

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]
/
sec2 x

cos2 x [(sec3 x) - (tan x - 1) (sec x tan x)]

sec x - cos2 x (sec x tan2 x - sec x tan x)

sec x - (cos x tan2 x + cos x tan x)

sec x - [(cos x tan x)(tan x + 1)]

10. Aug 26, 2013

### mileena

Thanks janhaa! I am not sure what you did though to get rid of the initial "1 -".

11. Aug 26, 2013

### haruspex

There's a sign error in that step.
What does cos x tan x simplify to?
Do you know sec2 = 1 + tan2 ? (Can derive this easily from cos2+sin2=1.)

12. Aug 26, 2013

### mileena

Ok, I see my error. I become very flustered with trig for some reason. I guess I get anxious when the pressure is on.

Ok, since tan x = sin x/cos x,
cos x tan x reduces to just sin x.

Here it is corrected:

sec x - cos2 x (sec x tan2 x - sec x tan x)

sec x - cos x tan2 x + cos x tan x

sec x - cos x tan2 x + sin x

And yes, I know that sec2 = 1 + tan2 is one of the three Pythagorean identities, but I don't see any sec2 or 1's above.

13. Aug 26, 2013

### haruspex

No, but there's a tan2.

14. Aug 26, 2013

### mileena

Oh my gosh. I think I have it!

sec x - cos x tan2 x + sin x

sec x - cos x (sec2 x -1) + sin x

sec x - cos x sec2 x - cos x + sin x

sec x -sec x - cos x + sin x

sin x - cos x

I hope it's that easy.

For some reason, I am not seeing all the basic trigonometric equivalents.