In △ABC, sides a, b, c are in A.P and 2/{1!9!} + 2/{3!7!} + 1/{5!5!} = 8a/(2b)! then the maximum value of tanA tanB is

Question

In △ABC, sides a, b, c are in A.P and \(\frac 2{1!9!} + \frac 2{3!7!} + \frac 1{5!5!} = \frac{8^a}{(2b)!}\) then the maximum value of tanA tanB is 

(a) 1/2

(b) 1/3

(c) 1/4

(d) 1/5

Answer 1

Correct option is (b) 1/3