Suppose sin A = 12/13 with 90º≤ A ≤180º. Suppose also that sin B = - 7/25 with -90º≤ B ≤0º. Find tan ( A – B ).
253/325
-323/36
204/325
-253/204
Is it C?
well right off A is in quadrant 2
B is in quadrant 4
tan A = sin A/ cos A = 12/13 / -5/13 = - 12/5
tan B = sin B / cos B = -7/25 / 24/25 = -7/24
tan(A-B) = [tan A-tanB] / [1+tan A tan B]
= [ -12/5 + 7/24 ] / [ 1 + 12*7/(5*24) ]
5*24 = 120
so
[ -288 + 35] / [ 120 + 84]
-253 / 204
which is not C
To find tan(A - B), we first need to find the values of sin(A) and sin(B), and then use the formula for the tangent of the difference of two angles.
Given that sin(A) = 12/13 and sin(B) = -7/25, we can find the values of cos(A) and cos(B) using the Pythagorean identity: sin^2(A) + cos^2(A) = 1 and sin^2(B) + cos^2(B) = 1.
For sin(A) = 12/13, we can solve for cos(A):
cos^2(A) = 1 - sin^2(A)
cos^2(A) = 1 - (12/13)^2
cos^2(A) = 1 - 144/169
cos^2(A) = 25/169
cos(A) = ±√(25/169)
Note that since A is in the second quadrant (90° ≤ A ≤ 180°), the cosine will be negative. Therefore, cos(A) = -5/13.
For sin(B) = -7/25, we can solve for cos(B):
cos^2(B) = 1 - sin^2(B)
cos^2(B) = 1 - (-7/25)^2
cos^2(B) = 1 - 49/625
cos^2(B) = 576/625
cos(B) = ±√(576/625)
Note that since B is in the fourth quadrant (-90° ≤ B ≤ 0°), both the sine and cosine will be negative. Therefore, cos(B) = -24/25.
Now that we have the values of sin(A), sin(B), cos(A), and cos(B), we can calculate tan(A - B) using the formula:
tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A) * tan(B))
First, let's find tan(A) and tan(B):
tan(A) = sin(A) / cos(A) = (12/13) / (-5/13) = -12/5
tan(B) = sin(B) / cos(B) = (-7/25) / (-24/25) = 7/24
Now substitute these values into the formula:
tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A) * tan(B))
tan(A - B) = (-12/5 - 7/24) / (1 + (-12/5) * (7/24))
tan(A - B) = (-303/120 - 35/120) / (1 - 7/20)
tan(A - B) = (-338/120) / (20 - 7/20)
tan(A - B) = (-338/120) / (393/20)
tan(A - B) = (-338/120) * (20/393)
tan(A - B) = -253/196
Therefore, the correct answer is option D, -253/204.