In a triangle tanA + tanB + tanC = 6 and tanA tanB = 2, then the values of tanA, tanB and tanC are respectively
Text Solution
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tan A + tan B + tan C = 6, tan A tan B = 2
In any Δ ABC,
tan A + tan B + tan C = tan A tan B tan C
⇒ 6 = 2 tan C ⇒ tan C = 3
∴ tan A + tan B + 3 = 6
⇒ tan A + tan B = 3 & tan A tan B = 2
Now (tan A – tan B)2 = (tan A + tan B)2 – 4tan A tan B
=9 – 8
= 1
⇒ tan A – tan B = ± 1
∴ tan A – tan B = 1 or tan A – tan B = – 1
tan A + tan B = 3 tan A + tan B = 3
on solving on solving
tan A = 2 tan A = 1
tan B =1 tan B = 2
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