Simplifying a trigonometry expression

[mileena]

Homework Statement

Simplify:

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

Homework Equations

none

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.

 

[haruspex]

[(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

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)?

 

[HallsofIvy]

Homework Statement

Simplify:

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

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 $sec^3(x)$?

Homework Equations

none

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 ] /

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 $(2/3)\sqrt{3}$.

 

[HallsofIvy]

Homework Statement

Simplify:

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

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 $sec^3(x)$?

Homework Equations

none

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 ] /

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.

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 $(2/3)\sqrt{3}$.

 

[mileena]

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)?

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.

 

[janhaa]

Thanks haruspex, and sorry for my late reply. I have a hard time getting online in the evenings.
1 - [[(tan² x - tan x)]/sec² x]
That's it. I don't know how to reduce this more.

maybe you can simplify it this way:

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

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

 

[mileena]

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 $sec^3(x)$?

Thanks for your reply HallsofIvy!

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

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

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

 

[haruspex]

(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)]

There's a mistake in how you multiplied out the (tan x - 1)(sec x tan x) term.

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

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.

 

[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)]

 

[mileena]

maybe you can simplify it this way:

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

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

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

 

[haruspex]

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

sec x - (cos x tan² x + cos x tan x)

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.)

 

[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.

 

[haruspex]

sec x - cos x tan² x + sin x
I don't see any sec² or 1's above.

No, but there's a tan².

 

[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.