## pvs285 one year ago (csc x + 1)(csc x - 1)

1. NoelGreco

FOIL it and apply trig identity

2. itsmylife

so its $(\csc x)^2-1=cot^22x$

3. NoelGreco

yep

4. pvs285

@itsmylife how did you get that

5. itsmylife

trig identity

6. pvs285

how

7. pvs285

can you show how you got that plz

8. itsmylife

pythagorean identity

9. NoelGreco

There are three Pythagorean trig identities that you simply must memorize.

10. itsmylife

go here @pvs285 http://www.sosmath.com/trig/Trig5/trig5/trig5.html

11. pvs285

can you just show the work to answer this question plz?

12. surjithayer

$\left( \csc x+1 \right)\left( \csc x-1 \right)=\csc ^{2}x-1=\cot ^{2}x$

13. itsmylife

$\sin^2x+\cos^2x=1,\frac{ \sin^2 x}{ \sin^2x }+\frac{ \cos^2x }{ \sin^2x }=\frac{ 1 }{ \sin^2x }, 1+\cot^2x=cosec^2x$

14. NoelGreco

It is against the policies of this site to simply provide answers.

15. pvs285

huh u said it was cot^2

16. pvs285

ik i whant you to explain it to me so i uderstand

17. itsmylife

it might be a mistake ;)

18. pvs285

huh

19. pvs285

pls help

20. pvs285

plz

21. surjithayer

$\csc ^{2}x-\cot ^{2}=1,or \csc ^{2}x-1=\cot ^{2}x$ It is an identity.

22. pvs285

so there is no work involved?

23. NoelGreco

pvs: it appears you are in over your head. If you have a question such as this, and can't pick up on the idea of a trig identity you might want to slow down and back up.

24. ybarrap

$$\tt \bf \Large(\csc x + 1)(\csc x -1)\\ \Large=\csc^2x-\csc x +\csc x -1\\ \Large=\csc^2x-1\\\\ \Large=\left(\dfrac{1}{\sin x}\right)^2-1\\ \Large=\dfrac{1}{\sin^2x}-1\\ \Large=\dfrac{1}{\sin^2x}-\dfrac{\sin^2x}{\sin^2x}\\\\ \Large=\dfrac{1-\sin^2x}{\cos^2x}\\\\ \Large=\dfrac{\cos^2x}{\sin^2x}\\\\ \Large=\cot^2x$$ Please let me kow if you have any questions. Good luck.