Proving more identites

Yes, that works too.  I always find it easier to remember just the sine and cosine relationships though, and work out the other trig identities from those if I need them.

.

Again, use sines and cosines

1/cos²x - 1/sin²x = sin²x/cos²x - cos²x/sin²x

right-hand side can be written as: (sinx/cosx + cosx/sinx)(sinx/cosx - cosx/sinx) 

                                              = (sin²x + cos²x)/(sinxcosx)*(sinx/cosx - cosx/sinx) 

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

                                              = 1/cos²x - 1/sin²x = right-hand side

NB:  You need to remember that sin²x + cos²x =  1.  This is an important identity.

Got it, thanks. I worked out i could have also used sec²x=1+tan²x and cosec²x=1+cot²x so that the 1s cancelled out.