Geometric justification of the trigonometric identity $\\arctan x + \\arctan \\frac{1-x}{1+x} = \\frac \\pi 4$?

Question

The trigonometic identity $$ \\arctan x + \\arctan \\frac 1 x = \\frac\\pi 2\\quad\ ext{for }x>0 $$ 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: 1. Use the usual formula for a sum of two arctangents; or 2. 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.

Answer 1

Yes. Consider the isosceles right triangle in the first quadrant with vertices (0,0), (x,1), and (x+1,1-x).

Draw a bounding rectangle with vertices (0,0), (x+1,0), (x+1,1), and (0,1).

Then using the definition of the tangent function applied to the three right triangles meeting at the origin gives the identity arctan x + arctan[(1-x)/(1+x)] = π/4