Question

which of the following is an identity?

A. csc^2x+cot^2x=1

B. (cscx+cotx)^2=1

C. sin^2x-cos^2x=1

D. sin^2x sec^2x+1=tan^2x csc^2x

Answer 1

Which one do you think?

Answer 2

wait

Answer 3

@magepker728 - Can you please refrain from giving users direct answers, as it hinders the learning process and is strongly prohibited according to the OS CoC. Thank you.

Answer 4

@Miracrown back off

Answer 5

Excuse me? I'm trying to enforce the rules here and you're evading it. Wow, that'll get you to places for sure!

Answer 6

@chalkzone - we don't know in advance
let's start with A and see if we can prove whether it is an identity or not
let's simplify the left side, and see what it makes.
convert it into terms of cos and sin

csc^2x + cot^2x = 1

Can you do that?

Answer 7

no not at all sadface

Answer 8

\[\frac{ 1 }{ \sin ^{2} x } \space + \frac{ \cos ^{2} x }{ \sin ^{2} x }\]

Answer 9

\[\frac{ 1+\cos ^{2}x }{ \sin ^{2}x }\]

\[\frac{ 1+\cos ^{2} x}{ 1-\cos ^{2} x }\]

is this an identity?

Answer 10

yes! I just looked at my notes

Answer 11

Hmm, well, I don't think so

Answer 12

in fact, I am quite certain there are only a few x values for which this is true

Answer 13

oh nevermind, the one I saw on my paper was vice versa

Answer 14

here is the deal: let's pretend it is an identity
then:

Answer 15

\[1+\cos^2x \space = 1-\cos^2x\]
\[2\cos^2x = 0\]
\[cosx = 0\]

Answer 16

is cos x = 0 for all x?

Answer 17

don't know

Answer 18

It should be no. Only specific x values make this true and identity is an equation that is true for ALL x... or, at least, true for all but maybe a few, like no dividing by zero, or something like that

Answer 19

this is not an identity.
let's look at b, and apply the same strategy

Answer 20

\[(cscx+cotx)^2 = 1\]

Answer 21

again, let's convert to sin and cos, like before ...

Answer 22

\[(\frac{ 1 }{ sinx } \space +\frac{ cosx }{ sinx } )^2 \space = 1\]
\[(\frac{ 1+cosx }{ sinx } )^2 \space = 1\]
\[\frac{ 1+2cosx+\cos^2x }{ \sin^2x } \space = 1\]

Answer 23

\[\frac{ 1+2cosx+\cos^2x }{ 1-\cos^2x } = 1\]

Answer 24

Predictions?

Answer 25

I am so clueless right now

Answer 26

\[1+2cosx+\cos^2x \space =1-\cos^2x\]
\[2cosx+2\cos^2x \space = 0\]
\[cosx(1+cosx) = 0\]
\[cosx= 0 , -1\]

Is this an identity?

Answer 27

Let me give a hand: Its not an identity
This may seem strange, but when we find the one that is an identity, you will see the difference

Answer 28

\[\sin^2x-\cos^2x=1\]
\[-\cos2x=1\]
\[\cos2x=-1\]

True for all x?

Answer 29

no

Answer 30

Right! :) So the answer must be D, but let's see why:

Answer 31

\[\sin^2xsec^2x+1 = \tan^2 x \csc^2x\]
\[\frac{ \sin^2x }{ \cos^2x } + 1 \space = \frac{ \sin^2x }{ \cos^2x } \space \frac{ 1 }{ \sin^2x }\]
\[\tan^2x+1 = \sec^2x\]
\[\sin^2x+\cos^2x=1\]

Answer 32

How about this?

Answer 33

oooooooo damn so that's what you need to do to get the equation

Answer 34

Yes! So now we see that this is an identity, because, using known identities, we reduce this to an expression where the left side is IDENTICAL to the right side.
so this is true, for ALL x hence why we call it an IDENTITY

Answer 35

I was so confused

Answer 36

thanks

Answer 37

yw :)