Question

Which of the following is NOT an identity?

A.1+cos^2x/sin^2x=1
B.(cosx + sinx)^2 = 1 + 2cosx sinx
C(tanx + cotx)^2 = csc^2 x + sec^2 x
D.tan^2x = sec^2x - sin^2x - cos^2x

Answer 1

Well, let's start with the first one and see what we can do with it.

\[\frac{ 1+\cos ^{2}x }{ \sin ^{2}x }\]
\[\frac{ 1 }{ \sin ^{2}x } + \frac{ \cos ^{2}x }{ \sin ^{2}x }\]
\[\csc ^{2}x + \cot ^{2}x = 1\]
Is this true? :P

Answer 2

no csc^2x - cot^2x=1

Answer 3

Well there ya go. Lucky us, first one was our answer.

Answer 4

okay thank you! :)

Answer 5

No problem ^_^

Answer 6

@Psymon can you help me with this one too please?

which of the following IS an identity

Answer 7

Yeah, ill take a look.

Answer 8

Personally, I would start with the problems that actually look like something we could use. C and D look the simplest to mess around with, so I'll start with D

Answer 9

\[\sin ^{2}xcot ^{2}x + \cos ^{2}xtan ^{2}x = 1\]
\[\frac{ \sin ^{2}xcos ^{2}x }{ \sin ^{2}x } + \frac{ \cos ^{2}xsin ^{2}x }{ \cos ^{2}x } = 1\]
\[\cos ^{2}x + \sin ^{2}x = 1\]
Lucky again, first one I tried :P. Really, you just gotta mess with the ones that actually look like something. I look at the first one and feel like I could mess with that forever and get nowhere, so I'd only do it as a last resort. 2nd one looks complex, too, so I just figured Id go in reverse order, do 4, 3, 2, 1 and hey, first one I tried :P

Answer 10

okay that makes sense, thank you so much! :D

Answer 11

Np :3