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

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

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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 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

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