Question

Prove this trig identity csc^4-csc^2=cot^4 cot^2

csc = 1/sin

csc^4 -csc^2 = cot^4 *cot^2

cot = cos/sin

1/sin^4 - 1/sin^2 = (cos/sin)^4 *(cos/sin)^2

(1 - sin^2)
-------------- = (cos^4)/(sin^4) *(cos^2)/(sin^2)
sin^4

sin^2 +cos^2 = 1 so from this result 1-sin^2 = cos^2

cos^2 cos^6
---------- = --------- cross multiply
sin^4 sin^6

sin^6 *cos^2 = sin^4 *cos^6 divide both sides by sin^4*cos^2

sin^2 = cos^4

is this far right ?

Answer 1

@TheSmartOne
@ganeshie8
@Kainui
@Nnesha

Answer 2

@zepdrix

Answer 3

@agent0smith

Answer 4

ty. Will.H

Answer 5

My pleasure

Answer 6

@Conqueror
@zpupster

Answer 7

@sweetburger

Answer 8

@.Sam.

Answer 9

@AloneS

Answer 10

\[\csc ^4x-\csc ^2x=\csc ^2x \left( \csc ^2x-1 \right)=\csc ^2x \cot ^2x\]

Answer 11

O_O

Answer 12

i think there is something wrong in the statement.

Answer 13

\[\csc ^2x \cot ^2x=\left( 1+\cot ^2x \right)\cot ^2x=\cot ^2x+\cot ^4x\]

Answer 14

I think, you don't need to calculate both sides
You can just calculating left or right side

Answer 15

yes Kevin i think it same you - bc. we need proving that the right or left side is equivalent with the other one

Answer 16

I will prove the left:
csc^4-csc^2 =
1/sin^4 - 1/sin^2 =
1/sin^4 - sin^2/sin^4 =
(1-sin^2)/sin^4

Answer 17

(1-sin^2)/sin^4 =
cos^2 / sin^4 =
(cos^4 + cos^2sin^2)/ sin^4 =
cos^4/sin^4 + (cos^2sin^2)/sin^4 =
cot^4 + cot^2

Answer 18

@Kevin but there in the first line what i posted above there are on the right side

cot^4 time cot^2 - so time not plus

Answer 19

what o_O

Answer 20

was your question correct?

Answer 21

i ve got it on the other math forum and i solved it till this step what you see on the top

Answer 22

@jhonyy9
You may want to verify your source.
See @sshayer's post above.
He is the first one to post a correct solution.

Answer 23

ok. ty.