If the value of 1 + tan 1 o 1 + tan 2 o 1 + tan 3 o … … … . ( 1 + tan 44 o ) ( 1 + tan 45 o ) is 2 λ , then the sum of the digits of the number λ is

Step-by-step solution

If, A + B = 45 o tan ⁡ A + B = 1 ∵ tan ( A + B ) = tan A + tan B 1 − tan A tan B ⇒ tan ⁡ A + tan ⁡ B = 1 - tan ⁡ A tan ⁡ B ⇒ 1 + tan A ( 1 + tan B ) = 2 LHS = 1 + tan 1 o ) ( 1 + tan 44 o ) ( 1 + tan 2 o ) 1 + tan 43 o ) … [ 1 + tan 45 o ) for each 1 + tan θ 1 + tan π 4 − θ = 2 = 2 22 1 + 1 = 2 23 = 2 λ then, λ = 23. Hence the sum of digits of λ is 2 + 3 = 5