We have to prove that
#cosalpha=sinbeta#, where
#alpha=2arctan(1/7)# and
#beta=4arctan(1/3)#.
So:
#tan(alpha/2)=1/7#
and
#tan(beta/4)=1/3#.
Remembering the double-angle formulae:
#sinx=(2t)/(1+t^2)#,
#cosx=(1-t^2)/(1+t^2)# where #t=tan(x/2)#
and
#sin2x=2sinxcosx#,
then:
#cosalpha=2sin(beta/2)cos(beta/2)rArr#
(where #t_1=tan(alpha/2)# and #t_2=tan(beta/4)#
#(1-t_1^2)/(1+t_1^2)=2*(2t_2)/(1+t_2^2)*(1-t_2^2)/(1+t_2^2)#
#(1-1/49)/(1+1/49)=2*(2*1/3)/(1+1/9)*(1-1/9)/(1+1/9)rArr#
#(48/49)/(50/49)=2*(2/3)/(9/10)*(8/9)/(9/10)rArr#
#48/49*49/50=2*2/3*9/10*8/9*9/10rArr#
#24/25=24/25#.