tan(A-B) = (tanA - tanB)\(1+tanA tanB) find tan 15 deg
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Here is your solution
Take A = 45° and B = 30°
Using the identity
tan(A-B) = (tanA - tanB) / (1+tanA tanB)
tan(45 -30) = (tan45 -tan30) / (1+ tan45*tan30)
tan15° =[ 1- (1/√3) ] / [ 1 + (1/√3)]
tan15 = (√3 -1) / (√3 +1)
on rationalization
tan15 = [ (√3 -1)^2] / 2
tan15 = (4 - 2√3) / 2
tan15 = 2 - √3
Hope this helps!!!!!
Here is your solution
Take A = 45° and B = 30°
Using the identity
tan(A-B) = (tanA - tanB) / (1+tanA tanB)
tan(45 -30) = (tan45 -tan30) / (1+ tan45*tan30)
tan15° =[ 1- (1/√3) ] / [ 1 + (1/√3)]
tan15 = (√3 -1) / (√3 +1)
on rationalization
tan15 = [ (√3 -1)^2] / 2
tan15 = (4 - 2√3) / 2
tan15 = 2 - √3
Hope this helps!!!!!
Yash0804:
Please ask if anything is not clear in the solution.
everything is clear.......thnx alt:)
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