anonymous
  • anonymous
Use basic identities to simplify the expression. cosecant of theta times cotangent of theta divided by secant of theta.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
write everything in terms of sine and cosine and see what cancels
anonymous
  • anonymous
\( \huge \frac{\csc \theta \times \cot \theta}{\sec \theta} \) This is what we are working with correct?
anonymous
  • anonymous
These things can get tricky but they are fun, when you are not being timed to do them lol

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anonymous
  • anonymous
Haha yeah and that is what the question is :)
DecentNabeel
  • DecentNabeel
@Nixy cot=cos/sin not sin/cos
anonymous
  • anonymous
opps yep thanks for the catch
anonymous
  • anonymous
\( \huge \frac{\frac{1}{\sin \theta} \times \frac{cos}{sin}}{\sec \theta} \) now work the top
anonymous
  • anonymous
Uhhh this is bad, I don't know how to....sorry I am taking this online and my teacher isn't that helpful :(
DecentNabeel
  • DecentNabeel
\[\mathrm{Simplify}\:\frac{\frac{1}{\sin \left(θ\right)}\cdot \frac{\sin \left(θ\right)}{\cos \left(θ\right)}}{\sec \left(θ\right)}:\quad 1\]
anonymous
  • anonymous
wait isn't it cos over sin?
anonymous
  • anonymous
Yes, he made a boo boo too :-D
DecentNabeel
  • DecentNabeel
yes
anonymous
  • anonymous
sec is 1/cos theta correct? so would I put that in?
DecentNabeel
  • DecentNabeel
yes
anonymous
  • anonymous
and then cancel out the cos?
anonymous
  • anonymous
@peachpi chime in any time to help :-) @peachpi is good at these. She/He has helped me on some before.
anonymous
  • anonymous
Oh okay thanks :)
anonymous
  • anonymous
you guys look like you have it under control
anonymous
  • anonymous
write the denominator as 1/cos Θ. then flip it to the reciprocal when you change division to multiplication |dw:1435872610978:dw|
anonymous
  • anonymous
nothing cancels but you can multiply straight across
anonymous
  • anonymous
so it comes out to cot theta ^2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
oh wow so I made it more complicated than it was
anonymous
  • anonymous
yeah that happens a lot with these. I've done one on here where I basically went around in circles twice before getting to the solution
anonymous
  • anonymous
Haha yeah oops my bad... anyways thank you so much to all of you!! I'll remember your usernames for sure because I'm sure I'll need help later on :)
anonymous
  • anonymous
I think I made a boo boo
anonymous
  • anonymous
yeah somewhere between the 5th and 6th steps
anonymous
  • anonymous
@peachpi
anonymous
  • anonymous
Nope still wrong... lol
LynFran
  • LynFran
how about \[\frac{ 1 }{ \sin \theta }*\frac{ \cos \Theta }{ \sin \Theta }/\sec\]\[\frac{ \cos \theta }{ \sin ^{2}\theta }*\cos\]\[\frac{ \cos ^{2}\theta }{ \sin ^{2}\theta } \]\[\cot ^{2}\theta \]
anonymous
  • anonymous
@LynFran is correct. @Nixy It looked like you were cancelling/dividing one part of the numerator but not the other
anonymous
  • anonymous
I got what LynFran got
anonymous
  • anonymous
\[ \huge \frac{\frac{1-sin^2}{sin^2 }}{1} \] \[ \huge \frac{\frac{cos^2}{sin^2 }}{1} \]
anonymous
  • anonymous
yeah that's it
anonymous
  • anonymous
\[ \huge \frac{\frac{1}{\sin \theta} \times \frac{cos}{sin}}{\sec \theta} \] \[ \huge \frac{\frac{cos}{\sin^2}}{\sec \theta} \] \[ \huge \frac{\frac{cos}{\sin^2}}{\frac{1}{cos}} \] \[ \huge \frac{\frac{cos}{\sin^2}}{\frac{1}{cos}} \times cos \] \[ \huge \frac{\frac{cos^2}{\sin^2}}{\frac{1}{cos}} \times cos \] \[ \huge \frac{\frac{cos^2}{\sin^2}}{1} \] \[ \huge \frac{cos^2}{sin^2} \] \[ \huge \cot^2 \]
anonymous
  • anonymous
These things are fun :-D

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