If tanA=1\\\\2 and tanB=1\\\\3 then cos2A=A sinBB sin2BC sin3B D none of these

tan A =1/2

then 

cosA =2/root(5)

Cos2A= 2cos^2(A) -1 => 2*4/5 – 1 => 3/5

 

Now from the options,

Given tanB = 1/3

sinB = 1/root(10)

cosB = 3/root(10)

sin2B = 2sinBcosB = > 6/10 = 3/5

Thus C option.

Cos2A in terms of tan =1-tan²A/1+tan²A  after putting value we get cos2A= 3/5 now sin2A in tan form, sin 2B= 2tanB/1+tan²B put value of tanB we get 3/5