Given that AC=4AD or AD=(AC)/4
Let angleDBC=theta_1 and angleABD=theta_2
Hence theta=theta_1-theta_2
taking tan both side
tantheta=tan(theta_1-theta_2)
tantheta=(tantheta_1-tantheta_2)/(1+tantheta_1 tantheta_2)
angle ABC=B=theta_1
in"Triangle" DAB
tantheta_2=(AD)/(AB)
replace AD by (AC)/4
tantheta_2=(AC)/(4AB)
but we know that in Triangle CAB
tantheta_1=tanB=(AC)/(AB)
So tantheta_2=(AC)/(4AB)=1/4tanB
Hence
tantheta=(tanB-tantheta_2)/(1+tanB tantheta_2)
tantheta=(tanB-1/4tanB)/(1+tanB 1/4tanB)=((4tanB-tanB)/4)/((4+tan^2B)/4)
4 is going to be cancel
tantheta=(3tanB)/((4+tan^2B)