## LynFran one year ago 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)

1. Loser66

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

2. LynFran

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

3. Loser66

yyyyyyyyyyyyyyyyyyyes

4. Loser66

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

5. Loser66

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

6. Loser66

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

7. Loser66

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

8. Loser66

got it?

9. LynFran

i think so thanks

10. Loser66

ok