A community for students.
Here's the question you clicked on:
 0 viewing
 2 years ago
Verify that this is an identity.
sec(beta)+csc(beta)
All over: 1+tan(beta)
is equal to=csc(beta)
 2 years ago
Verify that this is an identity. sec(beta)+csc(beta) All over: 1+tan(beta) is equal to=csc(beta)

This Question is Closed

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0hmm maybe rewrite tan B as \(\frac{\sin B}{\cos B}\)?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{sec\beta + csc\beta}{1+tan\beta} = \frac{\frac{sin\beta+cos\beta}{sin\beta cos\beta}}{\frac{sin\beta + cos\beta}{cos\beta}} = \frac{1}{sin\beta}=csc\beta\]

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1How can I make it bigger?

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\frac{1}{\cos B} + \frac{1}{\sin B}}{\frac{\cos B + \sin B}{\cos B}}\] \[\frac{\frac{\sin B + \cos B}{\sin B \cos B}}{\frac{\cos B + \sin B}{\cos B}}\] \[\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}\]

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0ugh @Callisto said it :P

GOODMAN
 2 years ago
Best ResponseYou've already chosen the best response.0No..wait..im still a little confused, how did you guys split it?

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large \frac{1}{\sin B} = \csc B\]

GOODMAN
 2 years ago
Best ResponseYou've already chosen the best response.0Yes..i know that, i know the basic identities.

GOODMAN
 2 years ago
Best ResponseYou've already chosen the best response.0I mean, like the second and third step you did.

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large \sec B + \csc B = \frac{1}{\cos B} + \frac{1}{\sin B}\] LCD \[\frac{\sin B + \cos B}{\sin B \cos B}\]

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{cos \beta} + \frac{1}{sin\beta} = \frac{1(sin\beta)}{cos \beta sin\beta} + \frac{1(cos \beta)}{sin\beta cos \beta} =\frac{sin\beta +cos\beta}{sin\beta cos\beta} \]

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0\[1 + \tan B = \frac{\cos B}{\cos B} + \frac{\sin B}{\cos B}\]

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sin B + \cos B}{\cos B}\]

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1\[1+tan\beta = 1+ \frac{sin\beta}{cos\beta} = \frac{cos\beta}{cos\beta} +\frac{sin\beta}{cos\beta} = \frac{sin\beta+cos \beta}{cos\beta} \]

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0the \(\sin B + \cos B\) cancels

GOODMAN
 2 years ago
Best ResponseYou've already chosen the best response.0Oh..okayy...basically i had to lay out the fractions? Arghh this stuff is weird.

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}\]\[\frac{\cos B}{\sin B \cos B}\]

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1When you make it like that, it's easier to simplify~ Maths is weird in nature... I think :S

GOODMAN
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, I think i got it from here, thanks :D
Ask your own question
Sign UpFind more explanations on OpenStudy
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.