Question

Use basic identities to simplify the expression.

cosine of theta to the second power divided by sine of theta to the second power. + csc θ sin θ

Answer 1

@Elsa213 I don't understand

Answer 2

@hartnn

Answer 3

@zepdrix

Answer 4

@Hero @IMStuck

Answer 5

Simba again!! :O
What's the monkey's name? I don't remember +_+

Answer 6

Rafikki lol any idea how to do this one? I have a list of identities but idk which to use

Answer 7

\(\Large\rm \frac{\cos^2\theta}{\sin^2\theta}+\csc\theta \sin\theta\)
this is the problem?

Answer 8

yep

Answer 9

Remember your cosecant identity?
\(\Large\rm \csc\theta=\frac{1}{\sin\theta}\)
That should help us simplify the second term.

Answer 10

Rafikki? Mm ok yah that sounds right :d
Scar was the cool one. so scury

Answer 11

could we make that part sin^2 by multiplying 1/sin and sin by sin?

Answer 12

I'm a personal fan of Kovu (scar's son in Lion King 2) :p

Answer 13

Oh ive never seen that one :o

Answer 14

Our second term we want to plug in our identity for csc(theta)

\(\Large\rm \color{orangered}{\csc\theta}\sin\theta=\color{orangered}{\frac{1}{\sin\theta}}\sin\theta\)
Multiplying,
\(\Large\rm \frac{\sin\theta}{\sin\theta}\)
and cancel stuff out,
you end up with 1 for your second term, yes?

Answer 15

For the first term, write it like this:
\(\Large\rm \left(\frac{\cos\theta}{\sin\theta}\right)^2\)
then recall your identity,
(cosx/sinx)=cotx

Answer 16

so we're left with just cotx?

Answer 17

Yes, but with a square on it,
\(\Large\rm \cot^2\theta+1\)
And we were left with a 1 from the second term

Answer 18

From there you can apply your Pythagorean Identity that involves cotangent and cosecant.

Answer 19

oh and cot^2+1=csc^2!

Answer 20

yay gj \c:/

Answer 21

got time for one more similar one?

Answer 22

sorry class ending :c i gotta pack up

Answer 23

aww okay, thanks for the help!