Question

please help me prove this identity im stuck not geting it out at all!
(1+tanx+cotx)(cosx-sinx)=(cscx/sec^2x -secx/csc^2x)

Answer 1

just distribute the left hand side, you can get the right hand side.

Answer 2

im ending up with \[(-\sin^2x/cosx+\cos^2x/sinx)\] when i distribute it

Answer 3

yyyyyyyyyyyyyyyyyyyes

Answer 4

or \(\dfrac{cos^2}{sin}-\dfrac{sin^2}{cos}\) right?

Answer 5

now, \(sin = \dfrac{1}{csc}\) hence \(csc = \dfrac{1}{sin}\) right?

Answer 6

and \(sec (x) =\dfrac{1}{cos(x)} \rightarrow cos(x) =\dfrac{1}{sec(x)}\)

Answer 7

hence the first term is \(\dfrac{csc (x)}{sec^2(x)}\)
do the same with the second term.
combine to get the right hand side

Answer 8

got it?

Answer 9

i think so thanks

Answer 10

ok