If (1+tanα)(1+tan4α)=2,α∈(0,π/16), and α is equal toπk then k is
(1+tanA)(1+tanB)=2⇒ tanA+tanB=1−tanAtanB⇒ tan(A+B)=1, i.e., A+B=π4 or α+4α=π4, i.e., α=π20