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^opps, show all steps too please.

\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out

I just don't understand how you went from sin(x+y)sec(x)sec(y) to that..

Yes, you have confused me..

now your job is to multiply
\[\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)\]\]

you get
\[\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)\] as a first step

okay.. then were the heck do you go?

okay. that makes some sense..

basically the same thing for the other side too then?

did you know \(\sec(x)=\frac{1}{\cos(x)}\)? this is really algebra at this step

yes, same thing for the second term

only this time you will get \(\tan(y)\)

Yes I did. Just didn't understand the problem..
so you get down to tan x=tanx?

ohh, yeah. tan x + tan y

that is it, yes

So is that your answer then? tan x + tan y?

that is my answer, yes

Okay, that makes sense. Do you have time to help me with one more problem?

sure

okay, let me take a screen shot.

I kind of understand it.. but not really :/

ok we can do this, but it is going to take a minute
ready?

Yes.

yes, we know sin u and cos v

we need \(cos(u)\) do you know how to find it?

Don't you have to make a triangle or something?

yes exactly

Okay, I'm not sure exactly how to come about finding it on a triangle

|dw:1362154017344:dw|

third side via pythagoras
\[\sqrt{61^2-11^2}=60\]

Wouldn't it be 50?

no
\[\cos(u)=\frac{60}{61}\]

adjacent over hypotenuse

Okay. so we still ned to find the sin one right?

right but actually i made a mistake
\[\cos(u)=-\frac{60}{61}\] because you are in quadrant 2

now we need \(\sin(v)\) and it is the same idea as before |dw:1362154367573:dw|

the third side is
\[\sqrt{41^2-40^2}=9\]so \(\sin(v)=\frac{9}{41}\)

now plug the numbers directly in to the formula

okay, then you just solve it right? multiplying and adding fractions basically?

\[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{61}\times(- \frac{60}{61})\]

yeah it is arithmetic from here on in
gotta run, good luck

thank you!

wait quick

oh i made a typo!!

for multiplying you flip the second fraction?

\[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{41}\times(- \frac{60}{61})\]

oh no

flip nothing
multiply straight across

okay. ill give it a try... thanks!

yw