Asked by Chelsie
Given that sin alpha=4/sqrt65, pi/2<alpha<pi, and cos beta=-3/sqrt13, tan beta>0, find the exact value of tan(alpha+beta)
Answers
Answered by
Reiny
I will use a for alpha and b for beta
sin a = 4/√65 , then cos a = -7/√65 in II
tana = sina/cosa = -4/7
cosb = -3/√13 , then sin b = -2/√13 in III
tanb = 2/3
tan(a+b) = (tana + tanb)/(1 - tanatanb)
= (-4/7 + 2/3)/(1 - (-4/7)(2/3))
= 2/29
sin a = 4/√65 , then cos a = -7/√65 in II
tana = sina/cosa = -4/7
cosb = -3/√13 , then sin b = -2/√13 in III
tanb = 2/3
tan(a+b) = (tana + tanb)/(1 - tanatanb)
= (-4/7 + 2/3)/(1 - (-4/7)(2/3))
= 2/29
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