Posted by NYSNC on Wednesday, December 7, 2011 at 3:59pm.
Actually, since tan(pi/2-x) = cot(x)
and cot = 1/tan it falls right out.
Or, using the sum of tangents formula,
tan(arctanx + arctan(1/x))
= [tan(arctan(x)) + tan(arctan(1/x))][1 - tan(arctan(x))*tan(arctan(1/x))]
= [x + 1/x]/[1 - x*1/x] = (x + 1/x)/0 = oo
tan pi/2 = oo
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