Question

=

Answer 1

\[\large \frac{\sin(\theta)}{1+\cos(\theta)}+\frac{1+\cos(\theta)}{\sin(\theta)} = 2\csc(\theta)\]


\[\large \frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{1+\cos(\theta)}{\sin(\theta)} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{1+\cos(\theta)}{\sin(\theta)} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{(1+\cos(\theta))(1+\cos(\theta))}{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{1+2\cos(\theta) + \cos^2(\theta)}{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


I'll let you finish up

Answer 2

I really dont understand the end factoring, I know the answer is 2/sin theta, but how the hell do you get that.

Answer 3

what do you mean by "the end factoring"

Answer 4

and yes, you're supposed to get the right side, so the final answer is practically given to you

Answer 5

The last step you wrote, how is that simplified down? Can you like cross it out or something im lost and have a quiz tomorrow :/

Answer 6

in that last step, I FOILED out (1+cos(theta))(1+cos(theta)) to get 1 + 2*cos(theta) + cos^2(theta)

Answer 7

once you had that, then?

Answer 8

then combine the fractions and see what simplifies

Answer 9

so multiply the top and bottom of sin theta/1+cos theta by sin theta, then multiply 1+cos theta/sin theta by (1+cos(theta)). Then foil that out,(1+cos(theta))(1+cos(theta)), which gives 1 + 2*cos(theta) + cos^2(theta). Then combine the fractions, simplify it down, and get 2/sin theta.?

Answer 10

yeah there's a bit more in between, but you have the right idea

Answer 11

Im horrible at fractions, so can you show those steps? the cos's simplify out?

Answer 12

ok hold on

Answer 13

\[\large \frac{\sin(\theta)}{1+\cos(\theta)}+\frac{1+\cos(\theta)}{\sin(\theta)} = 2\csc(\theta)\]


\[\large \frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{1+\cos(\theta)}{\sin(\theta)} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{1+\cos(\theta)}{\sin(\theta)} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{(1+\cos(\theta))(1+\cos(\theta))}{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))}+\frac{1+2\cos(\theta) + \cos^2(\theta)}{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{\sin^2(\theta)+1+2\cos(\theta) + \cos^2(\theta)}{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{(\sin^2(\theta)+\cos^2(\theta))+1+2\cos(\theta)}{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{(1)+1+2\cos(\theta) }{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{1+1+2\cos(\theta) }{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{2+2\cos(\theta) }{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


\[\large \frac{2(1+\cos(\theta)) }{\sin(\theta)(1+\cos(\theta))} = 2\csc(\theta)\]


I'm sure you see what to do from here

Answer 14

cross out so left with sin theta= 2csc theta, simplify that to 2sin theta?

Answer 15

then 2/sin(theta) turns into 2csc(theta)

Answer 16

ty

Answer 17

yw