Question

Trigonometry help! Proving Identities!

Answer 1

\[\sec^4\theta-\tan^4\theta=\frac{ 1+\sin^2\theta }{ \cos^2\theta }\]

Answer 2

do you know pythgorean identity in terms of sec and tan ?

Answer 3

Yes

Answer 4

by the way, the first step will be to use, \(a^2-b^2= (a+b)(a-b)\)
so, \(\sec^4 -\tan^4 = (\sec^2)^2-(\tan^2 )^2=.....?\)
good! what is it ?

Answer 5

\[1+\tan^2\theta=\sec^2\theta\]

Answer 6

so, from that, you get \(\sec^2-\tan^2=1\)
right ?

Answer 7

sorry, i am lazy enough to write theta everytime :P

Answer 8

Lol, it's fine. I thought that for the sec^4 -tan^4 that you have to foil them out or do you not?

Answer 9

applying this : \(a^2-b^2= (a+b)(a-b)\)
to sec^4 - tan^4
you get
\(\sec^4 -\tan^4 = (\sec^2)^2-(\tan^2 )^2=(\sec^2+\tan^2)(\sec^2-\tan^2)\)
makes sense ?

Answer 10

Yes

Answer 11

That's not the way I learned it in class lol.

Answer 12

okk...so that simplifies to ... ?

Answer 13

1/cos^2 + sin^2/cos^2

Answer 14

right!
and the next step ?

Answer 15

hint : notice the common denominator

Answer 16

1+sin^2/cos^2

Answer 17

so, you got it, right ?
any doubts anywhere ?

Answer 18

But.. I'm a bit confused. Does the sec^2-tan^2 cancel out?

Answer 19

it doesn't cancel out, it = 1
by using pythagorean identity
and (sec^2+tan^2) *1 = sec^2+tan^2

Answer 20

Oh okay, I got it. I meant to say it equals 1. Thanks!

Answer 21

welcome ^_^