Question

-tan^2x+sec^2x=1

Answer 1

If we're proving this identity we need to start with the following: \[\large \sin^2x+\cos^2x=1\]

we also have the following information: \[\large tanx=\frac{sinx}{cosx}\]

Answer 2

Okay

Answer 3

What would happened if divided the whole expression by say sin^2x or cos^2x?

Answer 4

Well I know that cos x= 1/sec x
But would that apply to this problem?

Answer 5

http://www.gradeamathhelp.com/image-files/trigonometric-identities-pythagorean.gif <--- notice the 2nd identity, solve for "1"

Answer 6

that works too

Answer 7

What would I do to start off? I would start on the left, right?

Answer 8

if you're solving for 1, you're going to have to get the tan^2 to the other side

Answer 9

okay, so it would look like this : sec^2x=1+tan^2x?

Answer 10

what would look like that? That's the identity you need to prove the original problem.

Answer 11

You said to move tan^2 to the other side, correct?

Answer 12

Oh, you're working with the original equation instead of the given 1+tan^2 = sec^2

Answer 13

My equation was -tan^2x+sec^2x=1.

Answer 14

Sigh. You need to use an identity you know is true before you can manipulate your equation.

You KNOW that 1+tan^2x= sec^2x

You use that to prove your equation

Answer 15

I know that but I don't know how to prove it

Answer 16

there isn't much to it.
\[1+\tan^2x= \sec^2x\]
\[1= \sec^2x - \tan^2x = - \tan^2x+\sec^2x\]

Answer 17

OH !!ok!! that's all I have to do?

Answer 18

as far as I can tell

Answer 19

THANK YOU!!

Answer 20

♥