anonymous
  • anonymous
Verify that this is an identity. sec(beta)+csc(beta) All over: 1+tan(beta) is equal to=csc(beta)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1337050670476:dw|
lgbasallote
  • lgbasallote
hmm maybe rewrite tan B as \(\frac{\sin B}{\cos B}\)?
Callisto
  • Callisto
\[\frac{sec\beta + csc\beta}{1+tan\beta} = \frac{\frac{sin\beta+cos\beta}{sin\beta cos\beta}}{\frac{sin\beta + cos\beta}{cos\beta}} = \frac{1}{sin\beta}=csc\beta\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.