Here's the question you clicked on:
skybunnygirl
(1+cos3t)/(sin3t)+(sin3t)/(1+cos3t)=2csc3t
\[\frac{ 1+\cos3t }{ \sin3t }+\frac{ \sin3t }{ 1+\cos3t }= 2\csc3t\] what do you want? prove it?
yes please, im having the hardest time with these questions...
I give you idea, you do the rest. the first part \[\frac{ 1+\cos3t }{ \sin3t }= \frac{ 1 }{ \sin3t }+ \frac{ \cos3t }{ \sin3t }= \csc 3t + \cot 3t\]
ok let me look at it
no go, have no idea what is next
That is pretty much it. You know what cot(x) is equal to? Also what is csc(x) equal to.
what else for cotx? there should be one more
now look at the first step, do you see something similar to what you said
sorry you lost me...
you said cotx = cosx/sinx and you said cscx=1/sinx right?
yes and that is what we converted into part of the original problem
ok I give you mine : from yours , right side = 2/ sin3t. write it out
now I just prove that left side * sin 2t =2, right? it is easy to do
\[\sin 3t \left[ \frac{ 1 + cost 3t }{ \sin3t } +\frac{ \sin3t }{ 1 + \cos 3t } \right]\]
\[= \frac{ \sin 3t + \cos 3t }{ \sin3t }+ \frac{ \sin^2 3t }{ 1 + \cos 3t }\]
ok, how did we get from csc3t+cot3t+sin3t/1+cos3t to what you writing now?
\[= (\frac{ sint }{ \sin3t }+ \frac{ \sin3tcos3t }{ \sin3t }+ \frac{ 1-\cos^2 3t }{1+\cos 3t })\]
\[=1+ \cos 3t + \frac{ (1+\cos3t)(1-\cos3t) }{ 1+\cos 3t }\]
and then you have 1+1 =2. the proof done. is it clear?
I make the right side2 csc 3t = 2/ sin3t. and then time both sides to sin 3t
do you get mine? think of it. write out and it will make sense to you.
ok let me look at it and see, im still not sure why we did the very first step as we ended up working out the original set up
our goal is prove the original equation is true, that means you must prove the left side = right side. some equation is not true at all, for example : how can you prove 1+1 =3? fortunately, your problem is prove it true, that means it is true already, just explain in which logic, it is true. done
ok, let me work it out on paper and see what I come up with. Thanks for your help.