I need complete help.. Complete the identity. Cos(a -b)/cos a cos b A. 1 + cot a cot b B. Tan a tan b + cot b C. 1 + tan a tan b D. 1 + cot a tan b

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I need complete help.. Complete the identity. Cos(a -b)/cos a cos b A. 1 + cot a cot b B. Tan a tan b + cot b C. 1 + tan a tan b D. 1 + cot a tan b

Mathematics
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try expanding the cos(a-b) part
How? I look at my textbook, but it isn't helping me understand more.
then separate the fraction after the expansion do some canceling and finally just a little rewriting

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Other answers:

you don't know the difference identity for cosine?
No. My textbook never showed or told me.
It just gives me problems to do and then I take my test.
sin(x-y)=sin(x)cos(y)-sin(y)cos(x) sin(x+y)=sin(x)cos(y)+sin(y)cos(x) cos(x+y)=cos(x)cos(y)-sin(x)sin(y) cos(x-y)=cos(x)cos(y)+sin(x)sin(y) so you never seen these identities? that is odd
Correct. I am working hard to learn.. But I am confused.
yes?
Can you help me a little better?
can you tell me if you applied that identity yet?
Like, I am confused. I have done some sins, cos, and tans but not like this.
do you see the cos(a-b) on top?
do any of the identities I mentioned look like this on the left hand side?
I do and yes.
well can you apply that identity then...
like can you expand cos(a-b)
The expanding is the confusing part. Never done it.
so you don't see how cos(x-y) looks exactly like cos(a-b) except in the place of x we have a and in the place of y we have b?
I see now.
replace the x and y on the right hand side with a and b respectively
and you have expanded cos(a-b)
Okay. Is that all I do?
to expand cos(a-b) yes for the question you have asked here at the top, no
I actually gave a few more steps above after the expansion
Okay. Thanks!
something about separating the fraction canceling stuff and rewriting a bit
can you show me what you have after rewriting \[\frac{\cos(a-b)}{\cos(a)\cos(b)}\]
or show me what you have for the numerator please
Yes
Cos(x-y)= cos(x) cos (y) I am still somewhat confused
ok well it looks like you didn't use the identity I mentioned above
cos(x-y)=cos(x)cos(y)+sin(x)sin(y)
cos(a-b)=?
|dw:1444190564321:dw|
Uhhh, ??? Sorry.. I have no clue. I have never done a problem like this. Cos(x-y)
|dw:1444190659448:dw|
|dw:1444190692534:dw|
Cos(x-y)= cos(x)cos(y) + sin(x)sin(y)
\[\frac{\cos(a-b)}{\cos(a)\cos(b)}=\frac{\cos(a)\cos(b)+\sin(a)\sin(b)}{\cos(a)\cos(b)}\] write as two fractions (separate the fraction)
Okay. I am understanding a little more.
So, 1 + cot a????
\[\frac{\cos(a) \cos(b)}{\cos(a) \cos(b)}=1\] and \[\frac{\sin(a) \sin(b)}{\cos(a) \cos(b)}=\frac{\sin(a)}{\cos(a)} \frac{\sin(b)}{\cos(b)}=...?\]
It equals... I have no clue. I am trying.
I thought it was 1
so you don't know what tan(x) equals in terms of sin(x) and cos(x)?
I do.. Sort of.
then what does it equal?
Tan a tan b?
tan(x)=tan(a)tan(b)?
or did you mean sin(a)/cos(a)=tan(a) and sin(b)/cos(b)=tan(b)?
I meant that
ok great \[\frac{\cos(a) \cos(b)}{\cos(a) \cos(b)}=1\] while \[\frac{\sin(a) \sin(b)}{\cos(a) \cos(b)}=\frac{\sin(a)}{\cos(a)} \frac{\sin(b)}{\cos(b)}=\tan(a) \tan(b)\]
I am on the right track?
we should be done
so I hope so :p
So, it is c ?
*?
yes that is what we got 1+tan(a)tan(b)
Yay! Thank you, thank you!

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