Like Cos(A)+Cos(B)=2 Cos[(A+B)/2].Cos [(A-B)/2] ,what is the formula for Tan x
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TanA-TanB
=Tan(A-B)(1-TanATanB)
= (sinAcosB-sinBcosA)/(cosAcosB)
= 2sin(A-B)/[cos(A - B) + cos(A + B)]
=Tan(A-B)(1-TanATanB)
= (sinAcosB-sinBcosA)/(cosAcosB)
= 2sin(A-B)/[cos(A - B) + cos(A + B)]
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tan(a) - tan(b)
tan(a) = sin(a) / cos(a)
tan(b) = sin(b) / cos(b)
sin(a) / cos(a) - sin(b) / cos(b)
[sin(a) cos(b) - sin(b) cos(a) ] / [cos(a) cos(b)]
But now note that the numerator is exactly sin(a - b)
[sin(a - b)] / [cos(a) cos(b)]
Now you can leave it like that, or change the 1/cos(a) and 1/cos(b) to secants:
sec(a) sec(b) sin(a - b)
It isn't really going to get a whole lot better than that.
Done!
P.S. tan(a) + tan(b) will just be sec(a) sec(b) sin(a + b)
tan(a) = sin(a) / cos(a)
tan(b) = sin(b) / cos(b)
sin(a) / cos(a) - sin(b) / cos(b)
[sin(a) cos(b) - sin(b) cos(a) ] / [cos(a) cos(b)]
But now note that the numerator is exactly sin(a - b)
[sin(a - b)] / [cos(a) cos(b)]
Now you can leave it like that, or change the 1/cos(a) and 1/cos(b) to secants:
sec(a) sec(b) sin(a - b)
It isn't really going to get a whole lot better than that.
Done!
P.S. tan(a) + tan(b) will just be sec(a) sec(b) sin(a + b)