Geometric justification of the trigonometric identity $\arctan x + \arctan \frac{1-x}{1+x} = \frac \pi 4$ The trigonometic identity
$$
\arctan x + \arctan \frac 1 x = \frac\pi 2\quad\text{for }x>0
$$
can be seen to be true by observing that if the lengths of the legs of a right triangle are $x$ and $1$, then the two acute angles of the triangle are the two arctangents above.
Does the identity
$$
\arctan x + \arctan \frac{1-x}{1+x} = \frac \pi 4
$$
have a similar geometric justification?
PS: The second identity, like the first, can be established by either of two familiar methods:


*

*Use the usual formula for a sum of two arctangents; or

*differentiate the sum with respect to $x$.
PPS: Secondary question: Both of the functions $x\mapsto\dfrac 1 x $ and $x\mapsto\dfrac{1-x}{1+x}$ are involutions.  Does that have anything to do with this?  Presumably it should mean we should hope for some geometric symmetry, so that $x$ and $\displaystyle\vphantom{\frac\int\int}\frac{1-x}{1+x}$ play symmetrical roles.
PPPS: In a triangle with a $135^\circ$ angle, if the tangent of one of the acute angles is $x$, then the tangent of the other acute angle is $\displaystyle\vphantom{\frac\int\int}\dfrac{1-x}{1+x}$.  That follows from the second identity above.  If there's a simple way to prove that result by geometry without the identity above, then that should do it.

Yes. Taking tan on both sides, 
$$ \frac{x + \frac{1-x}{1+x}}{1- x \frac{1-x}{1+x}
} = 1 $$ 
$$ \rightarrow \tan ^{-1} 1 = \pi/4 $$
EDIT1:
This is not a geometry based answer, but an analysis based, also am not too sure.
If  $$F(x) + F(1/x)= 2 c , $$
then,  $$ F(x) = F(1/x) = c.  \tag{1} $$
Proof by contradiction is that if it is not so,the Taylor expansion of F(x) would lead to singularities of F(1/x) for real  variable expansion in 1/x.
So given that 
$$  \tan^{-1} x + \tan^{-1} (1/x) = \pi/2   \tag{2} $$
Adding/subtracting same particular angle that may be included in geometric construction  $ (+\beta - \beta)  $ ( This had been done by Blue and David Quinn in a geometric proof)
$$ \tan^{-1} x +     \tan^{-1}\frac {1-x} {1+x}  -  \tan^{-1} \frac {1-x} {1+x}  +  \tan^{-1} (1/x) = \pi/2 $$
$$ (\tan^{-1} x +    \tan^{-1}\frac {1-x} {1+x} )   + ( \tan^{-1} (1/x) +   \tan^{-1} \frac{1-1/x} {1+1/x})  = \pi/2  \tag{3}  $$
which by virtue of (1) gives
$$ (\tan^{-1} x +    \tan^{-1} \frac{1-x} {1+x} ) =  \pi/4.  \tag{4} $$
