SOLUTION: In the 3rd quadrant , if sin A = -3/5 and sec B = -13/12 then tan (A-B) = ?

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Question 884194: In the 3rd quadrant , if sin A = -3/5 and sec B = -13/12 then tan (A-B) = ?
Answer by lwsshak3(11628) (Show Source):
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In the 3rd quadrant , if sin A = -3/5 and sec B = -13/12 then tan (A-B) = ?
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sinA=-3/5(working with a (3-4-5) reference right triangle in quadrant III)
cosA=-4/5
..
cosB=1/secB
cosB=-12/13(working with a (5-12-13) reference right triangle in quadrant III)
sinB=-5/13
..
sin(A-B)=sinA*cosB-cosA*sinB=(-3/5)*(-12/13)-(-4/5)*(-5/13)=36/65-20/65=16/65
cos(A-B)=cosA*cosB+sinA*sinB=(-4/5)*(-12/13)+(-3/5)*(-5/13)=48/65+15/65=63/65
tan(A-B)=sin(A-B)/cos(A-B)=16/63
Check:
sinA=-3/5
A≈216.87˚
cosB=-12/13
B≈202.62˚
A-B≈14.25
tan(A-B)≈tan(14.25)≈0.2539�
exact value=16/63≈0.2539�