If tan(A + B) = √3 and tan(A - B) = 1/√3; 0° < A + B ≤ 90° ; A > B, find A and B. Pls answer it correctly..​

Sorry don't know

45° & 15°

Step-by-step explanation:

=> tan(A + B) = √3 => tan(A + B) = tan60°

=> A + B = 60° ...(1)

=> tan(A - B) = 1/√3 => tan(A - B) = tan30°

=> A - B = 30° ...(2)

Adding (1) & (2):

=> (A + B) + (A - B) = 60° + 30°

=> 2A = 90°

=> A = 45°

Thus, A + B = 60° => B = 60° - 45° = 15°

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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