Proof:

Given: 3tanA.tanB = 1

To prove: 2cos(A B)=cos(A-B)

Proof:

We know that,

tan(A B) = (tanA + tanB) / (1 - tanA.tanB) - Equation 1

tan(A-B) = (tanA - tanB) / (1 + tanA.tanB) - Equation 2

From Equation 1, we can write,

tan(A B) = (tanA + tanB) / (1 - 1/(3tanA.tanB))

tan(A B) = (3tanA.tanB(tanA + tanB)) / (3tanA.tanB - 1) - Equation 3

From Equation 2, we can write,

tan(A-B) = (tanA - tanB) / (1 + 1/(3tanA.tanB))

tan(A-B) = (3tanA.tanB(tanA - tanB)) / (3tanA.tanB + 1) - Equation 4

Now, we can write,

2cos(A B) = 2 * (cos^2 (A B) - sin^2 (A B))

2cos(A B) = 2 * ((1 - sin^2 (A B)) - sin^2 (A B))

2cos(A B) = 2 - 4sin^2 (A B) - Equation 5

Also, we know that,

sin(A B) = (tanA + tanB) / (1 + tanA.tanB)

sin(A B) = (3tanA.tanB(tanA + tanB)) / (3tanA.tanB + 1) - Equation 6

From Equation 3 and 6, we can write,

sin(A B) / cos(A B) = (tanA + tanB) / (3tanA.tanB(tanA + tanB))

sin(A B) / cos(A B) = 1 / (3tanA.tanB - 1)

3tanA.tanB - 1 = 1 / (sin(A B) / cos(A B))

3tanA.tanB = (1 + sin(A B) / cos(A B))

3tanA.tanB = (cos(A B) + sin(A B)) / cos(A B)

3tanA.tanB = (cos(A B) / cos(A B)) + (sin(A B) / cos(A B))

3tanA.tanB = 1 + tan(A B)

3tanA.tanB - tan(A B) = 1 - Equation 7

Similarly, from Equation 4 and 6, we can write,

sin(A-B) / cos(A-B) = (tanA - tanB) / (3tanA.tanB + 1)

sin(A-B) / cos(A-B) = 1 / (3tanA.tanB + 1)

3tanA.tanB + 1 = 1 / (sin(A-B) / cos(A-B))

3tanA.tanB = (1 - sin(A-B) / cos(A-B

Just break tanA and tanB in sin cos and find the relation between them.. and then solve what u have to proof using cos(a+b) and cos(a-b) formula and u will get the ans..