Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A - B).
sketch right-angled triangles to match
sinA = 12/13
and
tanB = -4/3
from the first , y=12, r=13, so x = -5
and tanA = -12/5
tan(A-b) = (tanA - tanB)/(1 + tanAtanB)
= (-12/5 + 4/3)/(1 + (-12/5)(-4/3))
= -16/63
4/3
To find tan(A - B), we first need to find the values of A and B.
Given that sin A = 12/13 and 90º ≤ A ≤ 180º, we can use the Pythagorean identity to find cos A:
cos^2 A + sin^2 A = 1
cos^2 A + (12/13)^2 = 1
cos^2 A = 1 - (12/13)^2
cos^2 A = 1 - 144/169
cos^2 A = (169 - 144)/169
cos^2 A = 25/169
cos A = √(25/169)
cos A = 5/13
Since 90º ≤ A ≤ 180º, cos A is negative. So, cos A = -5/13.
Now, let's find the angle B:
Given that tan B = -4/3 and 270º ≤ B ≤ 360º, we can use the tangent identity to find cos B:
cos B = 1 / √(tan^2 B + 1)
cos B = 1 / √((-4/3)^2 + 1)
cos B = 1 / √(16/9 + 1)
cos B = 1 / √(16/9 + 9/9)
cos B = 1 / √(25/9)
cos B = 1 / (5/3)
cos B = 3/5
Since 270º ≤ B ≤ 360º, cos B is positive. So, cos B = 3/5.
Now that we have the values of sin A, cos A, sin B, and cos B, we can easily find tan (A - B):
tan(A - B) = (tan A - tan B) / (1 + tan A * tan B)
tan(A - B) = (sin A / cos A - sin B / cos B) / (1 + (sin A / cos A) * (sin B / cos B))
tan(A - B) = ( (sin A * cos B - cos A * sin B) / (cos A * cos B) ) / ( (cos A * cos B + sin A * sin B) / (cos A * cos B) )
tan(A - B) = (sin A * cos B - cos A * sin B) / (cos A * cos B + sin A * sin B)
Replacing the values:
tan(A - B) = ( (12/13) * (3/5) - (-5/13) * (-4/3) ) / ( (-5/13) * (3/5) + (12/13) * (-4/3) )
tan(A - B) = ( (36/65) - (20/39) ) / ( (15/39) - (48/65) )
tan(A - B) = ( (936/1521) - (1300/1521) ) / ( (585/1521) - (1872/1521) )
tan(A - B) = ( -364/1521 ) / ( -1287/1521 )
tan(A - B) = 364/1287
tan(A - B) = 4/11
Therefore, tan(A - B) = 4/11.
To find the value of tan(A - B), you need to use the trigonometric identities to express it in terms of the given values of sin A and tan B. Here's how you can do it:
1. Start by finding the values of cos A and cos B:
- Since sin A = opposite/hypotenuse = 12/13, you can use the Pythagorean identity to find cos A:
cos A = sqrt(1 - sin^2 A) = sqrt(1 - (12/13)^2) = sqrt(1 - 144/169) = sqrt(169 - 144)/13 = 5/13
- Since tan B = opposite/adjacent = -4/3, you can use the Pythagorean identity to find cos B:
cos B = sqrt(1 - tan^2 B) = sqrt(1 - (-4/3)^2) = sqrt(1 - 16/9) = sqrt(9 - 16)/3 = sqrt(-7)/3 (Note: In this case, cos B is negative because B lies in the third quadrant where cosine is negative.)
2. Next, use the identity tan(A - B) = (tan A - tan B) / (1 + tan A * tan B) to find tan(A - B):
- Substitute the values of sin A, tan B, cos A, and cos B into the formula:
tan(A - B) = (sin A/cos A - tan B) / (1 + (sin A/cos A) * tan B)
tan(A - B) = (12/13 / 5/13 - (-4/3)) / (1 + (12/13 / 5/13) * (-4/3))
tan(A - B) = (12/5 + 4/3) / (1 - 48/65)
tan(A - B) = (36/15 + 20/15) / (17/15)
tan(A - B) = (56/15) / (17/15)
tan(A - B) = 56/17
So, tan(A - B) is equal to 56/17.