How do you write 1/(sin(theta)+1) without fractions?

[Bonniee]

I thought at first I could replace 1/sin(theta) with cosecant theta, but realized that doesn't work because it's added to 1 inside the denominator. I tried many different ways as well like trying to cancel the fraction but this is an expression, and there can not be a single fraction in the expression.
PLEASE HELP!!!!!!

 

[blamocur]

Have you tried multiplying both the numerator and the denominator by [imath]1-\sin \theta[/imath]?

 

[Bonniee]

Yes, but that creates: (sin(theta)-1)/(sin^2(theta)-1),
And there was no identity that seemed useful.

 

[BigBeachBanana]

I thought at first I could replace 1/sin(theta) with cosecant theta, but realized that doesn't work because it's added to 1 inside the denominator. I tried many different ways as well like trying to cancel the fraction but this is an expression, and there can not be a single fraction in the expression.
PLEASE HELP!!!!!!

You can turn it into a [imath]\csc(...)[/imath] function which doesn't involve fractions, but its argument does.
What are you trying to achieve by getting rid of the "fraction"?

 

[Dr.Peterson]

Yes, but that creates: (sin(theta)-1)/(sin^2(theta)-1),
And there was no identity that seemed useful.

But the denominator certainly can be simplified by a Pythagorean identity ... though I would have multiplied by [imath]1-\sin(\theta)[/imath] rather than [imath]\sin(\theta)-1[/imath], as suggested in #2, to make it more obvious.

Once you do that, you have two terms that you can express without fractions.

I presume "without fractions" is a teacher's way of specifying what counts as "simplified".

 

[blamocur]

Yes, but that creates: (sin(theta)-1)/(sin^2(theta)-1),
And there was no identity that seemed useful.

[imath]1-\sin^2\theta = ??[/imath]

 

[Steven G]

sin²θ + cos²θ =1
So sin²θ - 1 =?