Question

tanx-cotx/sinxcosx=sec^2x-csc^2x

Answer 1

what would you like to do with this?

Answer 2

prove the identity. I need to take one side and make it equal the other

Answer 3

Ok, first lets start with the left side:

tanx - cotx / sinxcosx

(tanx - cotx) (1/sinxcosx)

tanx can be rewritten as sinx/cosx, and cotx can be rewritten as cosx/sinx sooooo:

(sinx/cosx - cox/sinx)(1/sinxcosx)


now lets distribute the 1/sinxcosx:

(sinx/cosx - cosx/sinx) (1/sinxcosx)

sinx/cosxsinxcosx - cosx/ sinxsinxcosx)

the sinx cancels out on the left side

1/cosxcosx - cosx/sinxsinxcosx

and similarly the cosx cancels out on the right

1/cosxcosx - 1/sinxsinx

cosxcosx is just cos^2x and sinxsinx is sin^2x so:

1/cos^2x - 1/sin^2x

but wait! isn't 1/cosx just secx? yes it is! so lets go ahead and rewrite that

sec^2x - 1/sin^2x

and isn't 1/sinx just cscx? well it is, so we can rewrite that too:

sec^2x - csc^2x

tada!

Answer 4

wow I wish it was that easy for me thanks!

Answer 5

Just takes a lot of practice, you'll get there.