Final answer:
To find tan(A+B) with tan A = 5/12 and sin B = 40/41, we use the angle sum identity for tangent and the Pythagorean identity to find cos B, and consequently tan B. We then substitute the values into the sum identity to get tan(A+B) = -525/92.
Step-by-step explanation:
If tan A = 5/12 and sin B = 40/41, and both angles A and B are in Quadrant I, to find the value of tan(A+B), we can use the angle sum identity for tangent, which is:
tan(A + B) = (tan A + tan B) / (1 - tan A * tan B)
Since we know tan A, we need to find tan B. Because sin B is given, we can use the Pythagorean identity sin²(B) + cos²(B) = 1 to find cos B, and then tan B = sin B / cos B.
sin²(B) = (40/41)² => cos²(B) = 1 - (1600/1681) => cos²(B) = 81/1681 => cos B = 9/41 (In Quadrant I, cosine is positive)
Therefore, tan B = sin B / cos B = (40/41) / (9/41) = 40/9
We can now substitute tan A and tan B into the angle sum identity:
tan(A + B) = (5/12 + 40/9) / (1 - (5/12 * 40/9))
tan(A + B) = (5/12 * 9 + 40 * 12) / (12 * 9 - 5 * 40) / (108/108)
tan(A + B) = (45/108 + 480/108) / (108/108 - 200/108)
tan(A + B) = 525/108 / -92/108
tan(A + B) = -525/92
So, tan(A+B) is -525/92.