SOLUTION: Using tan&#952; = sin&#952;/cos&#952;, prove the addition formula for tangent, tan(A + B) = (tan A + tan B) / (1

Click here to see ALL problems on Trigonometry-basics

Question 195575: Using tanθ = sinθ/cosθ, prove the addition formula for tangent,
tan(A + B) = (tan A + tan B) / (1 - tan A � tan B)

Answer by arallie(162) (Show Source):
You can put this solution on YOUR website!
tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)]
Proof:
(1) tan(a + b) = sin(a + b)/cos(a + b)
(2) sin(a + b) = (sin a)(cos b) + (cos a)(sin b)
(3) cos(a + b) = (cos a)(cos b) - (sin a)(sin b)
(4) So that:
sin(a + b)/cos(a + b) = [(sin a)(cos b) + (cos a)(sin b)]/[(cos a)(cos b) - (sin a)(sin b)] =
(sin a)(cos b)/[(cos a)(cos b) - (sin a)(sin b)] + (cos a)(sin b)/[(cos a)(cos b) - (sin a)(sin b)]
(5) Multiplying both sides of the fractions by 1/(cos a)(cos b) gives us:
(tan a)/[1 - (tan a)(tan b)] + (tan b)/[1 - (tan a)(tan b)] =
= [tan a + tan b ]/[1 - (tan a)(tan b)]

SOLUTION: Using tan&#952; = sin&#952;/cos&#952;, prove the addition formula for tangent, tan(A + B) = (tan A + tan B) / (1 - tan A � tan B)

SOLUTION: Using tanθ = sinθ/cosθ, prove the addition formula for tangent, tan(A + B) = (tan A + tan B) / (1 - tan A � tan B)