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Simplify : Tan^(-1) {a.cosx - b.sinx/ b.cosx + a.sinx} If x € (-p/2 , +p/2)

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Alright, after much searching I found the identity you really want. I'm sure you found while fiddling around with the

(a*cos(x) - b*sin(x)) / (b*cos(x) + a*sin(x))

that there wasn't really an identity that made that look any nicer. That's when I started looking at identities that had to do with Tan^(-1) which I will from now on refer to as arctan. It is an alternate notation that is useful when writing in text like this, but it means inverse tangent just like Tan^(-1) does.

So upon scrounging I found this identity:

arctan(c) - arctan(d) = arctan((c - d) / (1 + cd))

from this wikipedia page:

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Now I grant you, it still looks like we are nowhere close where we want to be, but we do kind of have something that has a minus up top and a plus on the bottom. I decided to divide the top and bottom of this expression:

(a*cos(x) - b*sin(x)) / (b*cos(x) + a*sin(x))

by b*cox(x) so that I would end up with a 1 in the spot where there was a 1 in the identity. I ended up with this:

((a/b) - tan(x)) / (1 + (a/b)tan(x))

But look! If we say that the value c = (a/b) and d = tan(x) then we have something that fits the identity! Which means we then have:

arctan(a/b) - arctan(tan(x))

Which simplifies to

arctan(a/b) - x

Which is loads easier than the original problem.

Hope that helps! Let me know if you have questions.