Sin A =1/2 and Cos B=12/13 where A lies in 1 quadrant and B lies in 3 quadrant. Then find tan(a-b).
To find the value of tan(A - B), we need to use trigonometric identities.
First, let's find the values of sin(A) and cos(B) individually, using the given information.
We are given that sin(A) = 1/2. Since A lies in the first quadrant, both sine and cosine are positive.
sin(A) = 1/2
Since sin(A) = opposite/hypotenuse, we can represent it as a right triangle with opposite side = 1 and hypotenuse = 2.
Using the Pythagorean Theorem, we can find the adjacent side:
adjacent^2 + opposite^2 = hypotenuse^2
adjacent^2 + 1^2 = 2^2
adjacent^2 + 1 = 4
adjacent^2 = 3
adjacent = sqrt(3)
So, cos(A) = adjacent/hypotenuse = sqrt(3)/2.
Next, we are given that cos(B) = 12/13. Since B lies in the third quadrant, cosine is negative, and sine is positive.
cos(B) = 12/13
Since cos(B) = adjacent/hypotenuse, we can represent it as a right triangle with adjacent side = 12 and hypotenuse = 13.
Using the Pythagorean Theorem, we can find the opposite side:
opposite^2 + adjacent^2 = hypotenuse^2
opposite^2 + 12^2 = 13^2
opposite^2 + 144 = 169
opposite^2 = 169 - 144
opposite^2 = 25
opposite = 5
So, sin(B) = opposite/hypotenuse = 5/13.
Now, we can use the difference formula for tangent (tan(A - B)):
tan(A - B) = (tan(A) - tan(B))/(1 + tan(A)tan(B))
To get tan(A), we divide sin(A) by cos(A):
tan(A) = sin(A)/cos(A) = (1/2)/(sqrt(3)/2) = 1/sqrt(3) = sqrt(3)/3
To get tan(B), we divide sin(B) by cos(B):
tan(B) = sin(B)/cos(B) = (5/13)/(12/13) = 5/12
Now substituting the values into the formula:
tan(A - B) = (sqrt(3)/3 - 5/12)/(1 + (sqrt(3)/3)(5/12))
To simplify the expression further, we need to find a common denominator:
tan(A - B) = [(4sqrt(3) - 5sqrt(3))/12] / [1 + (5sqrt(3)/36)]
Combining like terms in the numerator:
tan(A - B) = -sqrt(3)/12 / (1 + 5sqrt(3)/36)
To simplify further, multiply the numerator and denominator by the conjugate of the denominator to rationalize the expression:
tan(A - B) = [-sqrt(3)/12 * (36/36)] / [(1 + 5sqrt(3)/36) * (36/36)]
tan(A - B) = (-36sqrt(3)/432) / (36/36 + 5sqrt(3)/36)
tan(A - B) = -sqrt(3)/12 / (1 + 5sqrt(3)/36)
Now, simplifying the expression:
tan(A - B) = (-sqrt(3)/12) / (36/36 + 5sqrt(3)/36)
tan(A - B) = (-sqrt(3)/12) / ((36 + 5sqrt(3))/36)
Finally, simplifying further:
tan(A - B) = (-sqrt(3)/12) * (36/(36 + 5sqrt(
Therefore, the value of tan(A - B) is ((-36sqrt(3)(36))/((12*(36 + 5sqrt(3))))
To find tan(A-B), we can use the identities:
tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A) * tan(B))
Given that sin(A) = 1/2 and cos(B) = 12/13, we can find the values of tan(A) and tan(B) in the following way:
First, let's find the value of cos(A) using the Pythagorean identity:
cos^2(A) = 1 - sin^2(A)
cos^2(A) = 1 - (1/2)^2
cos^2(A) = 1 - 1/4
cos^2(A) = 3/4
cos(A) = √(3/4)
cos(A) = √3/2
Next, we can find the value of tan(A) using the identity:
tan(A) = sin(A) / cos(A)
tan(A) = (1/2) / (√3/2)
tan(A) = 1/√3
tan(A) = √3/3
For tan(B), we can use the identity:
tan(B) = sin(B) / cos(B)
But since we do not have the value of sin(B), we need to find it using the Pythagorean identity:
sin^2(B) = 1 - cos^2(B)
sin^2(B) = 1 - (12/13)^2
sin^2(B) = 1 - 144/169
sin^2(B) = (169 - 144) / 169
sin^2(B) = 25 / 169
sin(B) = √(25/169)
sin(B) = 5/13
Now we can calculate tan(B):
tan(B) = sin(B) / cos(B)
tan(B) = (5/13) / (12/13)
tan(B) = 5/12
Finally, we can 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) = (√3/3 - 5/12) / (1 + (√3/3)(5/12))
tan(A - B) = (√3/3 - 5/12) / (1 + √3/12)
tan(A - B) = (√3/3 - 5/12) / (1/12 + √3/12)
tan(A - B) = (√3/3 - 5/12) / ((1 + √3)/12)
tan(A - B) = 12(√3/3 - 5/12) / (1 + √3)
tan(A - B) = 4√3 - 5(4/12) / (1 + √3)
tan(A - B) = 4√3 - 20/12 / (1 + √3)
tan(A - B) = (4√3 - 20) / (12 + 12√3)
tan(A - B) = (4√3 - 20) / (12(√3 + 1))
Therefore, tan(A - B) = (4√3 - 20) / (12√3 + 12).
error in data
if B is in quad III, then cosB must be negative
I will assume you meant:
cos B = -12/13
make sketches of triangles in the two quadrants
for sinA = 1/2, y = 1, r = 2
x^2 + 1^2 = 2^2
x = √3
so tanA = y/x = 1/√3
for cosB = -12/13 in III
x = -12 , r = 13
then in III, y = -5
tanB = -5/-12 = 5/12
tan(A-B) = (tanA - tanB)/( 1 + tanAtanB)
= ((1/√3 - 5/12)/(1 + (1/√3)(5/12)
= ( (12 - 5√3)/(12√3) ) / ( (12√3 + 5)/(12√3) )
= (12 - 5√3)/(12√3 + 5)