## anonymous 3 years ago (csc x + 1)(csc x - 1)

1. NoelGreco

FOIL it and apply trig identity

2. anonymous

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

3. NoelGreco

yep

4. anonymous

@itsmylife how did you get that

5. anonymous

trig identity

6. anonymous

how

7. anonymous

can you show how you got that plz

8. anonymous

pythagorean identity

9. NoelGreco

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. NoelGreco

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

15. anonymous

huh u said it was cot^2

16. anonymous

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

17. anonymous

it might be a mistake ;)

18. anonymous

huh

19. anonymous

pls help

20. anonymous

plz

21. anonymous

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

22. anonymous

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.

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