three squares

There is a rather obscure formula for the sum of two inverse tangents:

 

tan^-1(a) + tan^-1(b)  =  tan^-1(  (a + b / ( 1 - a·b ) )

 

tan^-1( 1/2 ) + tan^-1( 1/3 )  =  tan^-1(  ( 1/2 + 1/3 ) / ( 1 - ( 1/2 )·( 1/3 ) )

                                        =  tan^-1( (5/6) / (5/6) )

                                        =  tan^-1( 1 )

                                        =  pi/4

 

Adding this to  tan^-1( 1 )  gives a total of  pi/2  [or 90^o]

 

sum  =  tan^–1(1/1) + tan^–1(1/2) + tan^–1(1/3)  

 

sum  =  45.000^o + 25.565^o + 18.435^o  

 

sum  =  89.000^o 

_.

alpha     =  tan^-1(1/1)

beta       =  tan^-1(1/2)

gamma  =  tan^-1(1/3)  

 

The sum will be  pi/2.