Question

Verify the identity:
1-(sin²ß/(1-cos²ß))= -cosß

Answer 1

\[1-\frac{\sin^2(B)}{1-\cos^2(x)}=1-\frac{\sin^2(B)}{\sin^2(B)}=1-1=0 \neq -\cos(B) \text{ for all B}\]

Answer 2

oops that x=B by the way

Answer 3

Were not sovling for ß, its verifying that 1-(sin²ß/(1-cos²ß)) does equal -cosß
lol

Answer 4

i'm not solving

Answer 5

i showed that one side is 0

Answer 6

-cos(B) is not 0 for all B
it is sometimes 0 for some B

Answer 7

OPPS ;o cos shouldn't be squared...

Answer 8

In other words, you have posted the problem incorrectly,.

Answer 9

\[1-\frac{\sin^2(B)}{1-\cos(B)}=-\cos(B)?\]

Answer 10

yes

Answer 11

i would multiply top and bottom of that fraction by the conjugate of the bottom

Answer 12

\[1-\frac{\sin^2(B)}{1-\cos(B)} \cdot \frac{1+\cos(B)}{1+\cos(B)}=1-\frac{\sin^2(x)(1+\cos(B))}{1-\cos^2(B)}\]

Answer 13

do you see what to do now?

Answer 14

you can use an identity at the bottom and something will cancel

Answer 15

yes, thank you! :D I have one more if you don't mind..?

Answer 16

ok

Answer 17

that didn't work

Answer 18

i can't open it correctly

Answer 19

oh.. one sec then