Suppose sin A = 12/13 with 90º≤A≤180º. Suppose also that sin B = 7/25 with -90º≤B≤0º. Find tan (A - B).
Well, isn't it a great day to solve some math problems and tan your brain cells with knowledge! Let's roll up our sleeves and dive right into it.
We have the values for sin A and sin B, so let's find the corresponding cos values first.
To find cos A, we'll use the identity sin^2 A + cos^2 A = 1. Since sin A = 12/13, we can substitute and solve for cos A:
(12/13)^2 + cos^2 A = 1
144/169 + cos^2 A = 1
cos^2 A = 1 - 144/169
cos^2 A = 25/169
cos A = 5/13
Similarly, for cos B, we use the identity sin^2 B + cos^2 B = 1. Substituting sin B = 7/25:
(7/25)^2 + cos^2 B = 1
49/625 + cos^2 B = 1
cos^2 B = 1 - 49/625
cos^2 B = 576/625
cos B = 24/25
Now, let's find the value of sin (A - B) using the formula: sin (A - B) = sin A * cos B - cos A * sin B:
sin (A - B) = (12/13) * (24/25) - (5/13) * (7/25)
sin (A - B) = (288/325) - (35/325)
sin (A - B) = 253/325
Finally, we can find the tan value using the identity: tan (A - B) = sin (A - B) / cos (A - B):
tan (A - B) = (sin (A - B)) / (cos (A - B))
tan (A - B) = (253/325) / (sqrt((1 - (253/325))^2 + (0)^2))
tan (A - B) = (253/325) / (sqrt((72/325)^2 + (325/325)^2))
tan (A - B) = (253/325) / (sqrt((72/325)^2 + 1))
tan (A - B) = (253/325) / (sqrt(5184/105625 + 105625/105625))
tan (A - B) = (253/325) / (sqrt(157809/105625))
tan (A - B) = (253/325) / (sqrt(2128/1425))
tan (A - B) = (253/325) / (sqrt(2128)/sqrt(1425))
tan (A - B) = (253/325) / (46.1111/37.7946)
tan (A - B) = (253/325) * (37.7946/46.1111)
tan (A - B) = 0.5923
So, the value of tan (A - B) is approximately 0.5923.
To find tan(A - B), we can use the formula for the tangent of the difference of two angles:
tan(A - B) = (tan A - tan B) / (1 + tan A * tan B)
First, let's find the values of tan A and tan B.
Given that sin A = 12/13, we can use the Pythagorean identity to find 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 = (169 - 144) / 169
cos^2 A = 25/169
Taking the square root of both sides, we get:
cos A = ± √(25/169)
cos A = ± 5/13
Since A lies in the second quadrant (90º < A < 180º), cos A should be negative:
cos A = -5/13
Now, we can find tan A:
tan A = sin A / cos A
tan A = (12/13) / (-5/13)
tan A = -12/5
Similarly, given that sin B = 7/25, we can use the Pythagorean identity to find 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 = (625 - 49) / 625
cos^2 B = 576/625
Taking the square root of both sides, we get:
cos B = ± √(576/625)
cos B = ± 24/25
Since B lies in the fourth quadrant (-90º < B < 0º), cos B should be positive:
cos B = 24/25
Now, we can find tan B:
tan B = sin B / cos B
tan B = (7/25) / (24/25)
tan B = 7/24
Next, let's substitute the values of tan A and tan B into the formula for tan(A - B):
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) = (-288/120 - 35/120) / (1 + (-84/120))
tan(A - B) = (-323/120) / (36/120)
tan(A - B) = -323/36
So, tan(A - B) = -323/36.
To find the value of tan (A - B), we need to use the formula for the tangent of the difference of two angles:
tan (A - B) = (tan A - tan B) / (1 + tan A * tan B)
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:
cos A = √(1 - sin^2 A)
= √(1 - (12/13)^2)
= √(1 - 144/169)
= √(169/169 - 144/169)
= √(25/169)
= 5/13
cos B = √(1 - sin^2 B)
= √(1 - (7/25)^2)
= √(1 - 49/625)
= √(625/625 - 49/625)
= √(576/625)
= 24/25
Now, we can find the values of tan A and tan B using the identities:
tan A = sin A / cos A = (12/13) / (5/13) = 12/5
tan B = sin B / cos B = (7/25) / (24/25) = 7/24
Substituting the values of tan A and tan B into the formula for tan (A - B):
tan (A - B) = (tan A - tan B) / (1 + tan A * tan B)
= (12/5 - 7/24) / (1 + (12/5) * (7/24))
= (288/120 - 35/120) / (1 + (84/120))
= (253/120) / (204/120)
= 253/204
Therefore, tan (A - B) = 253/204.