Question 835019: Find the value of tan(a+B) if csc=13/5, tanB= 4/3, and both angles are located in quadrant I.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the value of tan(a+B) if csc(a)=13/5, tanB= 4/3, and both angles are located in quadrant I.
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tan(a+B) = [tan(a)+tan(B)]/[1-tan(a)tan(B)]
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csc = r/y, so r = 13 and y = 5
Then x = sqrt[13^2-5^2] = 12
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Therefore: tan(a) = y/x = 5/12
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Your Problem:
tan(a+B) = [(5/12)+(4/3)]/[1-(5/12)(4/3)] = (21/12)/(5/9) = 35/36
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Cheers,
Stan H.
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