Question

help proving:
csc^4x-cot^4x = csc^2x+cot^2x ?
how to prove it?

Answer 1

Write it out in terms of cosine and sines

Answer 2

\[\frac{ 1 }{ \sin ^{4}x }-\frac{ \cos ^{4}x }{ \sin ^{4}x }\]

Answer 3

is the lhs, and now the rhs is?

Answer 4

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

Answer 5

so set them equal to each other and clear out the denominators by multipling through by the corresponding trigonometric functions

Answer 6

(solving on paper)

Answer 7

\[\frac{ \sin^{4}x-\sin^{4}xcos^{4}x }{ \sin^{4}x } = \frac{ \sin^{2}x+\sin^{2}xcos^{2}x }{ \sin^{2}x }\]
???

Answer 8

Note that \[1+\cot^2x=\csc^2x\]and that:\[\csc^4x-\cot^4x=(\csc^2x-\cot^2x)(\csc^2x+\cot^2x)\]

Answer 9

yes I know those identities but how will I apply it? help.

Answer 10

If \(1+\cot^2x=\csc^2x\), then what is \(\csc^2x-\cot^2x\)?

Answer 11

1

Answer 12

nvm, go with joe's idea much neater lol

Answer 13

Your done now @ECEstudent9405

Answer 14

So:\[\csc^4x-\cot^4x=(\csc^2x-\cot^2x)(\csc^2x+\cot^2x)=\cdots\]

Answer 15

ok. I get it sir. :)

Answer 16

thanks. so it can be solved just like that. (_ _)"