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प्रश्न
If tan (A + B) = `sqrt3` and tan (A – B) = `1/sqrt3`; 0° < A + B ≤ 90°; A > B, find A and B.
उत्तर १
tan (A + B) = `sqrt(3)` = tan 60° and tan (A – B) = `1/sqrt(3)` = tan 30°
A + B = 60° ...(1)
A – B = 30° ...(2)
2A = 90°
⇒ A = 45°
Adding (1) and (2)
A + B = 60°
A – B = 30°
Subtract equation (2) from (1)
A + B = 60°
A – B = 30°
2B = 30°
⇒ B = 15°
Note: sin(A + B) = sin A cos B + cos A sin B
sin(A + B) ≠ sin A + sin B
उत्तर २
Here, tan (A – B) = `1/sqrt(3)`
⇒ tan (A – B) = tan 30° ...[∵ tan 30° = `1/sqrt(3)`]
⇒ (A – B) = 30° ...(i)
Also, tan (A + B) = `sqrt(3)`
⇒ tan (A + B) = tan 60° ...[∵ tan 60° = `sqrt(3)`]
⇒ A + B = 60° ...(ii)
Solving (i) and (ii), we get:
A = 45° and B = 15°
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