Question

In class i was asked to show my steps of how you would go from cos^2(x)(sec^2x-1) to cot(x). I have no idea how to do this! Can you please show a step by step instructions. (I'm a VERY visual learner!)

Answer 1

do you know Pythagorean identities?

Answer 2

@DanJS you could have left what you had

Answer 3

all good..

Answer 4

I have the Pythagorean identities on a paper that i can use in class, but to be completely honest, my teacher didn't teach them very well.

Answer 5

you can divide both sides by cos^2(theta) to get the identity in terms of tan and sec

Answer 6

which results in
\[1+\tan^2(\theta)=\sec^2(\theta)\]

Answer 7

this is the identity dan had up a sec ago

Answer 8

i was thinking about drawing secant, cosecant lines and such on there, but then no

Answer 9

you can replace sec^2(x) with 1+tan^2(x) in your expression
see what this gives you after doing this

Answer 10

if you want you can show us what you have so we know you have done this

Answer 11

sorry i an super lost, but if you can list all the steps on one post, i can usually figure things out.

Answer 12

Well the first step I gave as: replacing sec^2(x) with 1+tan^2(x)
You will see 1-1=0 and be left with cos^2(x)(tan^2(x))
see if you can finish from here

Answer 13

here is the geometry of each trig function, you can see the right triangle for sin^2+cos^2=1^2 inside the circle
the other right triangle from
tan^2 + 1^2 = sec^2 is the larger bottom right triangle pythagorean theorem

Answer 14

ok, I'm still having a hard time comprehending all this Trig stuff (it's after midnight where I'm at), but I REALLy appreciate you trying to help. I think that I'll ask my teacher tomorrow.

Answer 15

you need to show
\[\cos^2(x)*(\sec^2(x) - 1) = \cot(x)\]

?

Answer 16

yep!

Answer 17

i dont know if those are equivalent

Answer 18

Anyone else getting \[\sin ^2(x)\] instead of cot(x)

Answer 19

can you check the prob again and see if we have it correct

Answer 20

oops...totally mixed two questions up!! (sorry everybody!) it should read cos^2(x)(sec^2x-1) is sin^2)(x)

Answer 21

Look at the photo i sent, perhaps you can understand it better

Answer 22

yep, there it is

Answer 23

i forgot to add, \[\tan^2(x) = \frac{ \sin^2(x) }{ \cos^2(x) }\]

Answer 24

thanks for your work! (sorry a typed everything down wrong!) that's the exact answer i needed!!

Answer 25

heres the edited photo if you need it

Answer 26

This was perfect!!! Thanks a million!!! :)