Question

sinthetadivided by 1-csctheta-sintheta divided by 1+cscthetaa =-2tansquaredtheta

Answer 1

this looks fun..lets see
\[\large \frac{\frac{\sin x}{1-\csc x -\sin x}}{1+\csc x}\]
is that it?

Answer 2

um i cant see the set up as yet it only has the words like fract

Answer 3

oh yeah site is like that sometimes..try refreshing page or restarting browser

Answer 4

k ill refresh again

Answer 5

anyway its like a fraction in numerator correct all over (1+csc)

Answer 6

yes

Answer 7

im not getting -2tan^2 ...hmm
http://www.wolframalpha.com/input/?i=%28sin%28x%29%2F%281-csc%28x%29-sin%28x%29%29%29%2F%281%2Bcsc%28x%29%29
check that im starting with correct expression?

Answer 8

no its sin theta divided by 1-csctheta(left side) minus sintheta divided by 1+csctheta =-2tan^theta

Answer 9

well crap

Answer 10

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

Answer 11

ok basically combine the fractions into 1 fraction and then it simplifies out pretty easily

rememeber sin * csc = 1

Answer 12

yeah @Jhannybean you want to post solution, i spent too much time trying to figure the wrong one out

Answer 13

ok thanks

Answer 14

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

Answer 15

I'm getting 1 on my numerator :\ am i doing something wrong....

Answer 16

should get 2 on top ... sin cancels then 1 -(-1) = 2

Answer 17

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

Answer 18

** all the 2tan^2 (theta)'s are all supposed to be negative on the right hand side >_<

Answer 19

I get a negative as well.

Answer 20

\[\large \frac{2}{-\cot^2(\theta)} = 2\tan^2(\theta)\]\[\large -2\tan^2(\theta)= -2\tan^2(\theta)\]

Answer 21

But if you look at the original problem, it is supposed to equal -2tan^2 θ

Answer 22

That's what i said....all the 2 tan^2 θ's I have written in my problem solving are supposed to be negatives. I don't want to go back and rewrite my latex to fix that.... T_T