what is tan(A-B) if SinA=1/2, cosA>0, tanB=3/4 and sinB<0
To find the value of tan(A-B), we need to use trigonometric identities and the given information about sinA, cosA, tanB, and sinB.
First, let's find the values of cosA and sinB using the given information.
Since sinA = 1/2 and cosA > 0, we can determine the value of cosA using the Pythagorean identity: cos^2(A) + sin^2(A) = 1.
Given sinA = 1/2, we have sin^2(A) = (1/2)^2 = 1/4. Solving for cos^2(A), we have:
cos^2(A) = 1 - sin^2(A) = 1 - 1/4 = 3/4.
Since cosA > 0, we take the positive square root: cosA = sqrt(3/4) = sqrt(3)/2.
Next, we know that tanB = 3/4 and sinB < 0. tanB = sinB/cosB, so we can find cosB.
tanB = sinB/cosB
3/4 = sinB/cosB.
Since sinB < 0 and cosB = sqrt(1 - sin^2(B)), we can determine cosB using the given information about sinB.
sin^2(B) = (sinB)^2 = 1 - cos^2(B).
Given sinB < 0, we have sin^2(B) = (-sinB)^2 = 1 - cos^2(B).
Since sinB < 0, cosB = -sqrt(1 - sin^2(B)).
Now, let's substitute these values into the formula for tan(A-B):
tan(A-B) = (tanA - tanB) / (1 + tanA * tanB).
Substituting tanA = sinA / cosA and tanB = sinB / cosB, we have:
tan(A-B) = (sinA / cosA - sinB / cosB) / (1 + (sinA / cosA) * (sinB / cosB)).
Plugging in the known values:
tan(A-B) = ([1/2] / [sqrt(3)/2] - [sinB / (-sqrt(1 - sin^2(B)))]) / (1 + ([1/2] / [sqrt(3)/2]) * ([sinB / (-sqrt(1 - sin^2(B)))])).
(note: I substituted sinA = 1/2 and cosA = sqrt(3)/2)
Simplifying the expression further using the known values, we will get the final value of tan(A-B).
sinA = 1/2, so tanA = 1/√3
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
= (1/√3 - 3/4)/(1+(1/√3)(3/4))
= (25√3 - 48)/39