We can use the identity:
tan(A + B) = (tan A + tan B) / (1 – tan A tan B)
Let A = tan^-1[2cos(sin^-1(1/2))] and B = tan^-1(1/3).
Then we have:
tan A = 2cos(sin^-1(1/2)) = 2(√3 / 2) = √3 sin A = √3 / (sec A) = √3 / 2
tan B = 1/3 sin B = 1/√(1+(1/3)^2) = 3/√10 cos B = √(1 – sin^2 B) = √(1 – 9/10) = 1/√10
Using the identity above, we have:
tan(A + B) = (tan A + tan B) / (1 – tan A tan B) = (√3 + 1/3) / (1 – √3/3) = [(3√3 + 1) / 9] / [(3 – √3) / 3] = (3√3 + 1) / (3 – √3)
Now, we can simplify the expression for A + B using the formula:
tan(A + B) = (tan A + tan B) / (1 – tan A tan B)
Thus, we have:
tan(A + B) = (3√3 + 1) / (3 – √3)
Therefore, the value of the expression tan^-1 [2cos(sin^-1(1/2))] + tan^-1(1/3) is equal to the inverse tangent of (3√3 + 1) / (3 – √3), which can be simplified as follows:
tan^-1[(3√3 + 1) / (3 – √3)] = tan^-1[(3√3 + 1) / (3 – √3) * (3 + √3) / (3 + √3)] = tan^-1[(9 + 4√3) / 6] = tan^-1[(3 + 2√3) / 2]
