Question

sec^4x - tan^4x

= sec^2x + tan^2x

how

Answer 1

you want that to be simplified?

Answer 2

err i mean proofed?

Answer 3

idk my book just says to match the trig expression with some stuff and the back of the book says it goes with that :/

Answer 4

idk how they get that

Answer 5

what's the whole question exactly?

Answer 6

Remember how to deal with the difference of squares? :) They can be broken up as 2 factors, conjugates of one another.\[(a^2-b^2)=(a-b)(a+b)\]In our problem here, we have something that looks like,\[(a^4-b^4)=\left[(a^2)^2-(b^2)^2\right]\]Which we can write as,\[(a^2-b^2)(a^2+b^2)\]
It's a little tricky if you don't understand what's going on with these squares :D
Does any of that make sense? How I was able to split up the 4th powers?

Answer 7

Match the trig expression with one of the following.

cscx

sinxtanx

sec^2x

sec^2x+tan^2x

Answer 8

so should it look like this
(sec^2x - tan^2x (sec^2x + tan^2x)

Answer 9

Yes. Looks good.
And from there, knowing that you're trying to get an answer of (sec^2x+tan^2x), we just want to somehow show that the other term is just 1.

So we'll need to use a handy identity.

Answer 10

\[\large \tan^2x+1=\sec^2x\]This is an important formula to remember.
If you plug this in for the sec^2x in (sec^2x-tan^2x), what happens? :O

Answer 11

it becomes 1 ? :o

Answer 12

\[\large (\sec^2x-\tan^2x)=(\tan^2x+1-\tan^2x)=(1)\]Oh my! :O I think yer right! Giving us,\[\large (\sec^2x-\tan^2x)(\sec^2x+\tan^2x)=(1)(\sec^2x+\tan^2x)\]

Answer 13

(sec2x−tan2x)=(tan2x+1−tan2x)=(1)

i got lost here :(

Answer 14

We need to utilize this identity.
\[\large [\sec^2x]=[\tan^2x+1]\]

\[\large ([\sec^2x]-\tan^2x)\]I'm using a little box now, so maybe it'll be easier to see where I'm substituting.\[\large ([\tan^2x+1]-\tan^2x)\]

Answer 15

oh ok thats better so do i plug [[\sec2x]=[\tan2x+1]\] into

(sec2x−tan2x)(sec2x+tan2x)

Answer 16

omg wait

Answer 17

i'm having difficulties with writing the equations :/

[sec2x]=[tan2x+1]

should it be plugged into

(sec2x−tan2x) (sec2x+tan2x)

Answer 18

both sides?

Answer 19

We only want to plug it into the factor with SUBTRACTION, because it will result in a nice cancellation.

We certainly could plug it in for both secant terms, but we don't need to :d

Answer 20

how come we couldnt turn secant into cosine

Answer 21

We could... but then we'd have to change tangent into sin/cos. And then we'd have to combine the fractions, and change 1-sin^2 into cos^2.. and then we could divide the fractions ...

Using this identity is much much simpler than going the sines and cosines direction.

Answer 22

one of the people that helped me with these kinds of problems told me the best way is to go with turning them into sines and cosines -.-

Answer 23

theres so many different ways to solve these its so hard :(

Answer 24

I told you that actually :c earlier I think. lol
It doesn't ALWAYS help to do so.
It's a good way to approach when you're really unsure though.

Yah there are a lot of identities.
It's hard to figure out which ones to use sometimes.
There are 3 equations that are really important for dealing with squared terms.

\[\large \sin^2x+\cos^2x=1\]\[\large \tan^2x+1=\sec^2x\]\[\large \cot^2x+1=\csc^2x\]

Answer 25

Lmao yup i think it was u haha thats funny

so after i plug in ([tan2x+1]−tan2x) it cancels out?

Answer 26

Here's a quick example...\[\large \tan^2x+\cot^2x\]If you look at the 3 identies above, does any ONE identity contain both tangent and cotangent? No, correct? They are in different formulas.

So this is an example where we would convert to sines and cosines, since we don't have a nice identity that converts tangent to cotangent, or cotangent to tangent.

Hopefully that distinction helps a little bit :\

Answer 27

After you plug in, we can move things around a little bit, since it's just addition and subtraction,\[\large (\tan^2x-\tan^2x+1)\]Do you see any nice cancellations? :O

Answer 28

this might blow you a lil l..but how come we couldnt plug in in [tan2x]=[sec2x+1] :o

Answer 29

blow me a lil???? :O WHU?
Anyway... how come? becauseeee that's .... not.... a thing :) lolol

The identity is as follows,\[\large \tan^2x+1=\sec^2x\]If we subtract 1 from each side, we get an identity in terms of tangent.\[\large \tan^2x=\sec^2x-1\]So that +1 that you wrote isn't quite right <:O understand why?
I mean, ... yes we certainly could have plugged in sec^2x-1 in place of the tan^2x :) Just make sure you do it with the -1.

Answer 30

Lol omg sorry for my slang!!!

so basically i can also write sin2 + cot2 = 1 as
cot2-1 = sin2

and 1 + cot2=csc2 can be csc2 - 1 = cot2

and sec2-1 = tan

Answer 31

Yes, that is very important actually.. sometimes you want to change 1-sin^2 into cos^2.. or the other way around.. that type of stuff comes up a lot :) so make sure you understand how you can move the terms around in those 3 identities.

Answer 32

wow youre the best thanks alot <3 !

Answer 33

wait one last question

so after i canncelled out the subtraction side,

i still have (sec2x + tan2x)

the book says thats the answer

but lets say i didn't have the answer from the book with me..how do i know when i'm done ?

Answer 34

Well in this case, it was a multiple choice question wasn't it? :o

That's a pretty tough question. After you broke it up into the conjugate pairs (the 2 sets of square thingy's, from the 4th power thingy...) you might have noticed that your problem then looked kinda similar to one of the options.

So you say to yourself, "hmm im funna see if i can make it look like dat one!"
And, voila! we were able to.

But yah.. that's a tough question :) You don't always know which multiple choice answer you're heading towards.