Question 931633
Find the exact value under the given conditions.
cos a = 1/3 , 0 < a < pi/2. sin b= -1/2 , -pi/2 < b < 0
Find tan ( a + b)
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reference angle a is in quadrant I where sin>0, cos>0
reference angle b is in quadrant IV where sin<0, cos>0
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cosa=1/3
sina=&#8730;(1-cos^2(a))=&#8730;(1-(1/9))=&#8730;(8/9)=&#8730;8/3
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sinb=-1/2
cosb=&#8730;(1-sin^2(b))=&#8730;(1-(1/4))=&#8730;(3/4)=&#8730;3/2
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tana=&#8730;8/1=&#8730;8
tanb=-1/&#8730;3=-&#8730;3/3
tan(a+b)=(tana+tanb)/(1-tana*tanb)=(&#8730;8-&#8730;3/3)/(1-(&#8730;8*-&#8730;3/3))
=(3&#8730;8-&#8730;3)/3)/(3+(&#8730;24)/3=(3&#8730;8-&#8730;3)/(3+&#8730;24)