Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A + B).
make quick sketches of right-angled triangles for each of the two cases, and use Pythagoras to find the missing sides.
first triangle:
since sinA = 12/13 , y = 12 and r = 13
x^2 + y^2 = r^2
x^2 + 144 = 169
x^2 = 25
x = ±5, but we are in quad II, so x = -5
then tanA = - 12/5
2nd triangle:
since tanB = -4/3
tan(A-B) = (tanA - tanB)/(1 + tanAtanB)
= (-12/5 - (-4/3) )/( 1 + (-12/5)(-4/3)
= (-16/15) / (21/5)
= -16/63
Are you sure it's negative?
To find the value of tan (A + B), we need to remember the trigonometric identity for the tangent of the sum of two angles, which states that:
tan (A + B) = (tan A + tan B) / (1 - tan A * tan B)
Given that sin A = 12/13, we can use the Pythagorean identity to find cos A:
cos A = √(1 - sin^2 A) = √(1 - (12/13)^2) = √(1 - 144/169) = √(25/169) = 5/13
Since cosine is positive in the fourth quadrant (270º ≤ B ≤ 360º) and tangent is equal to sin / cos, we can find cos B:
cos B = 1 / √(1 + tan^2 B) = 1 / √(1 + (-4/3)^2) = 1 / √(1 + 16/9) = 1 / √(25/9) = 3/5
Now that we have the values of sin A, cos A, tan B, and cos B, we can substitute them into the formula for tan (A + B):
tan (A + B) = (tan A + tan B) / (1 - tan A * tan B)
= (sin A / cos A + tan B) / (1 - (sin A / cos A) * tan B)
= (12/13 / 5/13 + (-4/3)) / (1 - 12/13 * (-4/3))
= (12/5 - 4/3) / (1 + 16/13)
= [(36 - 20) / 15] / (29/13)
= (16/15) / (29/13)
= 16/15 * 13/29
= 208/435
Therefore, tan (A + B) = 208/435.