Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Verify that this is an identity. sec(beta)+csc(beta) All over: 1+tan(beta) is equal to=csc(beta)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

|dw:1337050670476:dw|
hmm maybe rewrite tan B as \(\frac{\sin B}{\cos B}\)?
\[\frac{sec\beta + csc\beta}{1+tan\beta} = \frac{\frac{sin\beta+cos\beta}{sin\beta cos\beta}}{\frac{sin\beta + cos\beta}{cos\beta}} = \frac{1}{sin\beta}=csc\beta\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

lol..thats soo small.
How can I make it bigger?
\[\frac{\frac{1}{\cos B} + \frac{1}{\sin B}}{\frac{\cos B + \sin B}{\cos B}}\] \[\frac{\frac{\sin B + \cos B}{\sin B \cos B}}{\frac{\cos B + \sin B}{\cos B}}\] \[\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}\]
ugh @Callisto said it :P
No..wait..im still a little confused, how did you guys split it?
\[\large \frac{1}{\sin B} = \csc B\]
split what?
Yes..i know that, i know the basic identities.
I mean, like the second and third step you did.
\[\large \sec B + \csc B = \frac{1}{\cos B} + \frac{1}{\sin B}\] LCD \[\frac{\sin B + \cos B}{\sin B \cos B}\]
\[\frac{1}{cos \beta} + \frac{1}{sin\beta} = \frac{1(sin\beta)}{cos \beta sin\beta} + \frac{1(cos \beta)}{sin\beta cos \beta} =\frac{sin\beta +cos\beta}{sin\beta cos\beta} \]
\[1 + \tan B = \frac{\cos B}{\cos B} + \frac{\sin B}{\cos B}\]
\[\frac{\sin B + \cos B}{\cos B}\]
\[1+tan\beta = 1+ \frac{sin\beta}{cos\beta} = \frac{cos\beta}{cos\beta} +\frac{sin\beta}{cos\beta} = \frac{sin\beta+cos \beta}{cos\beta} \]
the \(\sin B + \cos B\) cancels
Oh..okayy...basically i had to lay out the fractions? Arghh this stuff is weird.
\[\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}\]\[\frac{\cos B}{\sin B \cos B}\]
When you make it like that, it's easier to simplify~ Maths is weird in nature... I think :S
Okay, I think i got it from here, thanks :D

Not the answer you are looking for?

Search for more explanations.

Ask your own question