Solution:
Given, tan A = 1/2 and tan B = 1/3
Let us first find the values of sin A, cos A, sin B and cos B using the following formulae:
tan A = sin A/cos A and tan B = sin B/cos B

Finding sin A and cos A:
tan A = 1/2
Let x be the angle whose tangent is 1/2. Then we have:
tan x = 1/2
Using the inverse tangent function, we get:
x = tan^-1(1/2) ≈ 26.57°
Since tan A = tan x, we have:
A = x ≈ 26.57°
Using the Pythagorean identity, we have:
sin^2 A + cos^2 A = 1
sin^2 A + (1/4) = 1
sin^2 A = 3/4
sin A = ±√3/2
Since A is in the first quadrant, we have:
sin A = √3/2
cos A = ±1/2
Since A is in the first quadrant, we have:
cos A = 1/2

Finding sin B and cos B:
tan B = 1/3
Let y be the angle whose tangent is 1/3. Then we have:
tan y = 1/3
Using the inverse tangent function, we get:
y = tan^-1(1/3) ≈ 18.43°
Since tan B = tan y, we have:
B = y ≈ 18.43°
Using the Pythagorean identity, we have:
sin^2 B + cos^2 B = 1
sin^2 B + (9/4) = 1
sin^2 B = -5/4
Since sin B cannot be negative, we have no solution for sin B.
cos B = ±√13/3
Since B is in the first quadrant, we have:
cos B = √13/3

Using the double angle formula for cosine:
cos 2A = 2cos^2 A - 1
Substituting the value of cos A, we get:
cos 2A = 2(1/2)^2 - 1 = 0
Using the formula for sin 2B:
sin 2B = 2sin B cos B
Substituting the value of sin B and cos B, we get:
sin 2B = 2(√3/2)(√13/3) = √39/3

Therefore, the correct option is (B) sin 2B.

Cos2A = 1-tan²A/ 1+ tan²A = 3/5...
Now, sin 2B = 2tanA / 1+ tan²A = 3/5 ...
So, Cos2A = Sin2B...