Evaluate the expression under the given conditions.
tan(A+B); cosA=-1/7, A in quadrant 3, sinB=1/2, B in quadrant 2.
in Q3, tan is positive, so tanA = √48
in Q2, tan is negative, so tanB = -1/√3
Now apply the addition formula
tan(A+B) = (tanA+tanB)/(1-tanA tanB)
To evaluate the expression `tan(A+B)`, we need to find the values of `A` and `B` first.
Given that `cosA = -1/7` and `A` is in quadrant 3, we can use the Pythagorean trigonometric identity `sin^2(A) + cos^2(A) = 1` to find the value of `sinA`:
sin^2(A) + cos^2(A) = 1
sin^2(A) + (-1/7)^2 = 1
sin^2(A) + 1/49 = 1
sin^2(A) = 1 - 1/49
sin^2(A) = (49 - 1)/49
sin^2(A) = 48/49
Since `A` is in quadrant 3, `sinA` is negative. Hence,
sinA = -sqrt(48/49)
= -sqrt(48)/sqrt(49)
= -sqrt(48)/7
= -4sqrt(3)/7
Given that `sinB = 1/2` and `B` is in quadrant 2, we can use the Pythagorean trigonometric identity to find the value of `cosB`:
sin^2(B) + cos^2(B) = 1
(1/2)^2 + cos^2(B) = 1
1/4 + cos^2(B) = 1
cos^2(B) = 1 - 1/4
cos^2(B) = (4 - 1)/4
cos^2(B) = 3/4
Since `B` is in quadrant 2, `cosB` is negative. Hence,
cosB = -sqrt(3/4)
= -sqrt(3)/sqrt(4)
= -sqrt(3)/2
Now, to find `tan(A+B)`, we can use the trigonometric identity `tan(A+B) = (tanA + tanB) / (1 - tanA * tanB)`:
tan(A+B) = (tanA + tanB) / (1 - tanA * tanB)
= (sinA / cosA + sinB / cosB) / (1 - (sinA * sinB) / (cosA * cosB))
= (-4sqrt(3)/7 / (-1/7) + 1/2 / (-sqrt(3)/2)) / (1 - ((-4sqrt(3)/7) * (1/2)) / ((-1/7) * (-sqrt(3)/2)))
Now, simplify the expression inside the parentheses:
(-4sqrt(3)/7 / (-1/7) + 1/2 / (-sqrt(3)/2)) = (4sqrt(3)/1 + 1/1) = 4sqrt(3) + 1
Substituting this back into the original expression:
tan(A+B) = (4sqrt(3) + 1) / (1 - ((-4sqrt(3)/7) * (1/2)) / ((-1/7) * (-sqrt(3)/2)))
= (4sqrt(3) + 1) / (1 - (2sqrt(3)/7) / (sqrt(3)/14))
= (4sqrt(3) + 1) / (1 - (2sqrt(3) * 14)/(7sqrt(3)))
= (4sqrt(3) + 1) / (1 - (28sqrt(3))/(7sqrt(3)))
= (4sqrt(3) + 1) / (1 - 4)
= (4sqrt(3) + 1) / (-3)
= -(4sqrt(3) + 1) / 3
Therefore, the value of `tan(A+B)` under the given conditions is `-(4sqrt(3) + 1) / 3`.