in QII sinA = 3/5, tanA = -3/4
in QIII cosB = -24/25, tanA = 7/24
Now just apply tour sum formula
tan(B-A) = (tanB-tanA)/(1 + tanB tanA)
Given sin A = 3/5, cos B = -24/25, 90° < A < 180°, 180° < B < 270°
Find tan(B - A)
2 answers
sinA = 3/5, and A is in II
then cosA = -4/5
leads to tan A = (3/5) / (-4/5) = -3/4
cos B = -24/25 , B in III,
then sinB = -7/25
which lead to tan B = sinB/cosB = (-7/25) / (-24/25) = 7/24
I got the above two by recognizing the 3-4-5 and 7-24-25 right-angled triangles.
You could of course make sketches and use Pythagoras to find them.
tan (B-A) = (tanB - tanA)/(1 + tanAtanB)
= ((7/24) - (-3/4))/(1 + (-3/4)(7/24))
= (25/24) / ( 25/32)
= 4/3
then cosA = -4/5
leads to tan A = (3/5) / (-4/5) = -3/4
cos B = -24/25 , B in III,
then sinB = -7/25
which lead to tan B = sinB/cosB = (-7/25) / (-24/25) = 7/24
I got the above two by recognizing the 3-4-5 and 7-24-25 right-angled triangles.
You could of course make sketches and use Pythagoras to find them.
tan (B-A) = (tanB - tanA)/(1 + tanAtanB)
= ((7/24) - (-3/4))/(1 + (-3/4)(7/24))
= (25/24) / ( 25/32)
= 4/3