Question

Last question...

Answer 1

cot^2x - csc^2x = -1 for all values of x

Answer 2

is this supposed to be a proof?

Answer 3

No, just true or false. But I'd like to know how to solve it...

Answer 4

probabably best bet is to start with the mother of all trig identities, \[\sin^2(x)+\cos^2(x)=1\]

Answer 5

then think of how you can get a cotangent or a cosecant out of it
it is not too hard, do you know what the definition of cosecant is?

Answer 6

or cotangent for that matter, either way

Answer 7

csc^2 = 1/sin^2

Answer 8

And cot is cos^2/sin^2

Answer 9

right exactly
you notice that sine is in the denominator of both right?

Answer 10

so you will get them both if you start with \[\sin^2(x)+\cos^2(x)=1\] and divide every term by sine

Answer 11

(cos^2 - 1)/sin^2 = -1
Is this correct so far? sorry, I'm really tired..

Answer 12

divide each term by sine squared, give \[\frac{\sin^2(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1}{\sin^2(x)}\]

Answer 13

then a small amount of algebra gets you exactly what you want

Answer 14

if that is not clear, let me know

Answer 15

It looks false, right?

Answer 16

no it is true

Answer 17

:/ how is that? gosh I'm really tired... sorry

Answer 18

\[\frac{\sin^2(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1}{\sin^2(x)}\] is the same as \[1+\cot^2(x)=\csc^2(x)\]

Answer 19

one or two algebra steps gets you to \[\cos^2(x)-\csc^2(x)=-1\]

Answer 20

Oh wow - that's so obvious! *pace palm*
Thank you so much.

Answer 21

yw