Question

Simplifying complex trigonometric identities pls help again....

Answer 1

\[\sec^4u-\sec^2u-\tan^4u\]

Answer 2

Does the identity apply if the power is 4?

Answer 3

Maybe start by factoring out sec^2 from first two terms

Answer 4

I can do that? I thought it had to be from all the terms

Answer 5

You're free to do anything you want as long as you don't violate the rules

Answer 6

factoring the GCF from first two terms is okay

Answer 7

ok so it would be
\[\sec^2u(\sec^2-1)+\tan^4u\]

Answer 8

i got 0 is this right? :D

Answer 9

+tan^4 or -tan^4?

Answer 10

\[\sec^2u(\tan^2u)+\tan^4u\]

Answer 11

\[\sec^2u(\tan^2u)\color{red}{-}\tan^4u\]

Answer 12

oh yea my bad

Answer 13

so now i guess you multiple sec and tan?

Answer 14

multiply*

Answer 15

which wud be one right?

Answer 16

No, factor out tan^2 instead

Answer 17

factor it out? like this?
\[\sec^2\tan^2(1)-\tan^4\]

Answer 18

You have \[\sec^2u(\color{blue}{\tan^2u})\color{red}{-}\color{blue}{\tan^4u}\]

Answer 19

see tan^2 common ?

Answer 20

oh yea

Answer 21

so it would be
\[\sec^2u \tan^2u(1-\tan^2u)\]

Answer 22

Nope

Answer 23

\[\sec^2u(\color{blue}{\tan^2u})\color{red}{-}\color{blue}{\tan^4u}\]

\[\color{blue}{\tan^2u}[\sec^2u\color{red}{-}\color{blue}{\tan^2u}]\]

Answer 24

oh ok

Answer 25

doesnt that equal
\[\tan^2u(1)\]

Answer 26

Yep

Answer 27

Ok wow thanks a lot :D.

Answer 28

There is a another short way to do this

Answer 29

:o

Answer 30

simply rewrite sec and tan in terms of sin and cos

Answer 31

\[\sec^4u-\sec^2u-\tan^4u\]

\[\dfrac{1}{\cos^4u}-\dfrac{1}{\cos^2u}-\dfrac{\sin^4u}{\cos^4u}\]

Answer 32

get a common denominator and add up the fractions

Answer 33

give it a try when you're free

Answer 34

i actually tried this method at first

Answer 35

it works nicely

Answer 36

but i think i got lost. ill try it now

Answer 37

Okay good luck

Answer 38

\[\frac{ 1 }{ \cos^4 }-\frac{ \cos^2 }{ \cos^4 }- \frac{ \sin^4 }{ \cos^4 }\]

Answer 39

Did i get the denominator right?

Answer 40

\[\frac{ 1-\cos^2-\sin^4 }{ \cos^4 }\]

Answer 41

\[\frac{ \sin^2-\sin^4 }{ \cos^4 }\]

Answer 42

Yes factor out sin^2

Answer 43

\[\frac{ 1-\sin^2 }{ \cos^4 }\]

Answer 44

\[\frac{ -\cos^2 }{ \cos^4 }\]

Answer 45

\[\frac{ -\sin^2 }{ \cos^2 }\]

Answer 46

-tan^2?

Answer 47

did i do something wrong?

Answer 48

whats 1-sin^2 ?

Answer 49

-cos^2

Answer 50

?

Answer 51

Are you sure ?

Answer 52

isnt the identity sin^2+cos^2=1 ?

Answer 53

Yes

Answer 54

so ?

Answer 55

oh wow

Answer 56

i see it now its cos^2

Answer 57

good haha

Answer 58

silly me

Answer 59

then that makes it work :D wow ty