anonymous
  • anonymous
The expression csctheta-sinthetais equivalent to what?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
\[\large \csc \theta-\sin \theta\] Remember your identity for cosecant? You can rewrite it terms of sine.
anonymous
  • anonymous
i know csctheta is 1/sin theta but i dont know what to do
zepdrix
  • zepdrix
\[\large \frac{1}{\sin \theta}-\sin \theta\]Ok from here, combine the fractions. Get a common denominator.

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zepdrix
  • zepdrix
\[\large \frac{1}{\sin \theta}-\frac{\sin\theta}{1}\] Understand how to do that?
anonymous
  • anonymous
why did we flip the right side?/
zepdrix
  • zepdrix
I didn't flip it. I'm just try to show you that \(\large 5\) is the same thing as \(\large \dfrac{5}{1}\). Maybe that way it will be easier for you to realize that they're both fractions. And can be combined.
anonymous
  • anonymous
ok..and find the common denominator which is 1 and sin theta
zepdrix
  • zepdrix
yah that sounds right, so we need to multiply the right fraction by \(\large \dfrac{\sin\theta}{\sin\theta}\), while the left fraction needs to be multiplied by 1/1 (so we can leave it alone).
zepdrix
  • zepdrix
\[\large \frac{1}{\sin \theta}-\frac{\sin\theta}{1}\color{royalblue}{\left(\frac{\sin\theta}{\sin\theta}\right)} \qquad = \qquad \frac{1}{\sin\theta}-\frac{\sin^2\theta}{\sin\theta} \qquad = \qquad \frac{1-\sin^2\theta}{\sin\theta}\] Confused about any of that? The blue term is the one we used to get a common denominator.
anonymous
  • anonymous
would the answer be 1?
zepdrix
  • zepdrix
No :o
anonymous
  • anonymous
it has a)1/sin theta x b)1 c)sintheta/cos^2theta d)cottheta costheta
zepdrix
  • zepdrix
we haven't finished the problem yet..
zepdrix
  • zepdrix
was trying to see if you're following any of this before going further :o
anonymous
  • anonymous
ohh so sin theta divided by sintheta is the lcd and you multiply it bythe sintheta/1
zepdrix
  • zepdrix
ya :o
anonymous
  • anonymous
ok i c
zepdrix
  • zepdrix
\[\large \frac{1-\sin^2\theta}{\sin\theta}\] From here it's all about remembering identities. There are 2 identities that we'll apply to match one of the answers. Do you remember an identity for this? \(\large 1-\sin^2\theta=?\)
anonymous
  • anonymous
cos^2 theta
zepdrix
  • zepdrix
\[\large \frac{1-\sin^2\theta}{\sin\theta} \qquad = \qquad \frac{\cos^2\theta}{\sin\theta}\]Ok good! From here we're going to split up the cosines on top,\[\large \frac{\cos^2\theta}{\sin\theta}\qquad=\qquad \frac{\cos \theta\cdot \cos \theta}{\sin \theta} \qquad=\qquad \frac{\cos \theta}{\sin \theta}\cdot \cos \theta\]
zepdrix
  • zepdrix
And from here we need to remember another identity. \(\large \dfrac{\cos\theta}{\sin\theta}=?\)
anonymous
  • anonymous
cot theta

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