Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

NoelGreco Group TitleBest ResponseYou've already chosen the best response.0
FOIL it and apply trig identity
 11 months ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
so its \[(\csc x)^21=cot^22x\]
 11 months ago

pvs285 Group TitleBest ResponseYou've already chosen the best response.0
@itsmylife how did you get that
 11 months ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
trig identity
 11 months ago

pvs285 Group TitleBest ResponseYou've already chosen the best response.0
can you show how you got that plz
 11 months ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
pythagorean identity
 11 months ago

NoelGreco Group TitleBest ResponseYou've already chosen the best response.0
There are three Pythagorean trig identities that you simply must memorize.
 11 months ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
go here @pvs285 http://www.sosmath.com/trig/Trig5/trig5/trig5.html
 11 months ago

pvs285 Group TitleBest ResponseYou've already chosen the best response.0
can you just show the work to answer this question plz?
 11 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
\[\left( \csc x+1 \right)\left( \csc x1 \right)=\csc ^{2}x1=\cot ^{2}x\]
 11 months ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
\[\sin^2x+\cos^2x=1,\frac{ \sin^2 x}{ \sin^2x }+\frac{ \cos^2x }{ \sin^2x }=\frac{ 1 }{ \sin^2x }, 1+\cot^2x=cosec^2x\]
 11 months ago

NoelGreco Group TitleBest ResponseYou've already chosen the best response.0
It is against the policies of this site to simply provide answers.
 11 months ago

pvs285 Group TitleBest ResponseYou've already chosen the best response.0
huh u said it was cot^2
 11 months ago

pvs285 Group TitleBest ResponseYou've already chosen the best response.0
ik i whant you to explain it to me so i uderstand
 11 months ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
it might be a mistake ;)
 11 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
\[\csc ^{2}x\cot ^{2}=1,or \csc ^{2}x1=\cot ^{2}x\] It is an identity.
 11 months ago

pvs285 Group TitleBest ResponseYou've already chosen the best response.0
so there is no work involved?
 11 months ago

NoelGreco Group TitleBest ResponseYou've already chosen the best response.0
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.
 11 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
$$ \tt \bf \Large(\csc x + 1)(\csc x 1)\\ \Large=\csc^2x\csc x +\csc x 1\\ \Large=\csc^2x1\\\\ \Large=\left(\dfrac{1}{\sin x}\right)^21\\ \Large=\dfrac{1}{\sin^2x}1\\ \Large=\dfrac{1}{\sin^2x}\dfrac{\sin^2x}{\sin^2x}\\\\ \Large=\dfrac{1\sin^2x}{\cos^2x}\\\\ \Large=\dfrac{\cos^2x}{\sin^2x}\\\\ \Large=\cot^2x $$ Please let me kow if you have any questions. Good luck.
 11 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.