Use basic identities to simplify the expression.
cosecant of theta times cotangent of theta divided by secant of theta.

- anonymous

- chestercat

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- anonymous

write everything in terms of sine and cosine and see what cancels

- anonymous

\( \huge \frac{\csc \theta \times \cot \theta}{\sec \theta} \) This is what we are working with correct?

- anonymous

These things can get tricky but they are fun, when you are not being timed to do them lol

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## More answers

- anonymous

Haha yeah and that is what the question is :)

- DecentNabeel

@Nixy cot=cos/sin
not sin/cos

- anonymous

opps yep thanks for the catch

- anonymous

\( \huge \frac{\frac{1}{\sin \theta} \times \frac{cos}{sin}}{\sec \theta} \) now work the top

- 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

\[\mathrm{Simplify}\:\frac{\frac{1}{\sin \left(θ\right)}\cdot \frac{\sin \left(θ\right)}{\cos \left(θ\right)}}{\sec \left(θ\right)}:\quad 1\]

- anonymous

wait isn't it cos over sin?

- anonymous

Yes, he made a boo boo too :-D

- DecentNabeel

yes

- anonymous

sec is 1/cos theta correct? so would I put that in?

- DecentNabeel

yes

- anonymous

and then cancel out the cos?

- anonymous

Oh okay thanks :)

- anonymous

you guys look like you have it under control

- anonymous

write the denominator as 1/cos Θ.
then flip it to the reciprocal when you change division to multiplication
|dw:1435872610978:dw|

- anonymous

nothing cancels but you can multiply straight across

- anonymous

so it comes out to cot theta ^2?

- anonymous

yes

- anonymous

oh wow so I made it more complicated than it was

- 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

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

I think I made a boo boo

- anonymous

yeah somewhere between the 5th and 6th steps

- anonymous

- anonymous

Nope still wrong... lol

- 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

I got what LynFran got

- anonymous

\[ \huge \frac{\frac{1-sin^2}{sin^2 }}{1} \]
\[ \huge \frac{\frac{cos^2}{sin^2 }}{1} \]

- anonymous

yeah that's it

- 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

These things are fun :-D

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