Question

help in verifying identities:

Answer 1

\[\sin ^{2}x \tan ^{2}x +\sec ^{2}x=\sec ^{4}x\]

Answer 2

Hello Miss KatKat! c:

Do you remember how to convert tangent into sines and cosines?

Answer 3

Yep!

Answer 4

\[\Large\rm \sin ^{2}x \left(\color{orangered}{\tan^{2}x}\right) +\sec ^{2}x=\sec ^{4}x\]

Answer 5

So what can we do with that orange guy? :d

Answer 6

\[\sin ^{2}x \div \cos ^{2}x\]

Answer 7

\[\Large\rm \sin ^{2}x \left(\color{orangered}{\frac{\sin^2x}{\cos^2x}}\right) +\sec ^{2}x=\sec ^{4}x\]Mmm k good good good.
From there what can we do?

Oooo hold up, I misread the problem >.<
Lemme think about this one a sec.. not sure if this is the way to go.

Answer 8

haha how are you making those equations look like that?

also, this problem probably uses the power reducing formulas.

Answer 9

If you right-click my fancy math stuff,
`Show Math As` > `TeX Commands`
You can see the way I typed it in.
Gotta know a few fun little tricks to be able to do it hehe :D

Answer 10

Power reductions for secant?
Hmm I know the power reduction for secant using integrals.
And I also know the Half-Angle formula helps to reduce powers, but that messes up our angle x.

What formula you speaking of? :U

Answer 11

brb i need some food in my belly :U

Answer 12

Are you sure you copied the problem down correctly?
This one seems kinda crazy 0_o

Answer 13

shoot i didnt.
\[\Large\rm \sec ^{2}x \left(\color{magenta}{\tan^{2}x}\right) +\sec ^{2}x=\sec ^{4}x\]

Answer 14

thats it, i apologize

Answer 15

Still a 4th power though? :U

Answer 16

theses are those formulas i was talking about

Answer 17

yep

Answer 18

could this work?

Answer 19

one sec

Answer 20

Yah those are the Half-Angle Identities, but notice how the `u` changes to `2u`?
That would mess up our angle :) So we can't use those. darn.

Answer 21

Remember your Pythagorean Identity involving tangent and secant?
That one should help us :)

Answer 22

i feel like i could still use those?

could we split the 4th power into sec^{2}x squared?

Answer 23

Yes. But we don't need to.
What we want to do is recall our Pythagorean Identity for tangent/secant.\[\Large\rm \tan^2x+1=\sec^2x\]That one look familiar?
We want to solve it for tan^2x.

Answer 24

\[\Large\rm \sec ^{2}x \left(\color{orangered}{\sec^{2}x-1}\right) +\sec ^{2}x=\sec ^{4}x\]

Answer 25

oh i got it!

Answer 26

Ooo nice job \c:/
hehe getting the hang of those colors too huh?

Answer 27

\[\Large\rm \sec ^{4}x \left(\color{lime}{-\sec ^{2}x}\right) +\sec ^{2}x=\sec ^{4}x\]

ignore the parentheses, thats subtraction not multiplication

but they cancel and sec=sec?

Answer 28

Good, yes.
When you distribute the sec^2x to each term in the brackets,
you get THAT (without the brackets).

And then a nice cancellation! yay!

Answer 29

sec^4x=sec^4x

Answer 30

yep! yay thanks so much :)

Answer 31

You've probably done enough of these that you're used to the format by now.
But just in case,
we make sure that we `leave one side of the equation alone, and try to match the other side to it`.

So we can't do things like "add this to each side".
You probably already knew that :) just making sure though hehe

Answer 32

yeah haha its a good reminder! thanks again for your help, i really appreciate it.