How do you solve $A r c tan (1 - x) - a r c tan x = π$ $Arc tan(1-x) - arc tanx = pi$ ?

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Answer 1 · 2016-05-23T08:23:34

$x = \frac{1}{2}$ $x=1/2$

As $arctan (1 - x) - arctan x = π$ $arctan(1-x)-arctanx=pi$

$arctan (1 - x) = π + arctan x$ $arctan(1-x)=pi+arctanx$

Now taking tangent of both sides

$tan (arctan (1 - x)) = tan (π + arctan x) = tan (arctan x)$ $tan(arctan(1-x))=tan(pi+arctanx)=tan(arctanx)$

or $1 - x = x$ $1-x=x$

or $2 x = 1$ $2x=1$ or $x = \frac{1}{2}$ $x=1/2$

How do you solve Arc tan(1-x) - arc tanx = piArctan(1−x)−arctanx=πArc tan(1-x) - arc tanx = pi?