Question

(csc x + 1)(csc x - 1)

Answer 1

FOIL it and apply trig identity

Answer 2

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

Answer 3

yep

Answer 4

@itsmylife how did you get that

Answer 5

trig identity

Answer 6

how

Answer 7

can you show how you got that plz

Answer 8

pythagorean identity

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

huh u said it was cot^2

Answer 16

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

Answer 17

it might be a mistake ;)

Answer 18

huh

Answer 19

pls help

Answer 20

plz

Answer 21

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

Answer 22

so there is no work involved?

Answer 23

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.

Answer 24

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

Good luck.