prove 1-cosB divided by 1+cosB= tan^2B/2
To prove the equation (1 - cos(B))/(1 + cos(B)) = tan^2(B/2), we can use a trigonometric identity known as the Half-Angle Identity for Tangent.
The Half-Angle Identity for Tangent states that tan(B/2) = (1 - cos(B))/(sin(B)). Now, let's manipulate this identity to match our equation.
First, we divide both sides of the identity by sin(B):
tan(B/2) / sin(B) = (1 - cos(B))/(sin(B)) * (1/sin(B))
tan(B/2) / sin(B) = (1 - cos(B))/(sin^2(B))
Next, we can use another trigonometric identity: sin^2(B) + cos^2(B) = 1. By rewriting sin^2(B) as 1 - cos^2(B), we get:
tan(B/2) / sin(B) = (1 - cos(B))/((1 - cos^2(B)))
Now, we can rewrite the denominator by factoring it:
tan(B/2) / sin(B) = (1 - cos(B))/((1 - cos(B))(1 + cos(B)))
Notice that the (1 - cos(B)) terms cancel out in the numerator and denominator:
tan(B/2) / sin(B) = 1/(1 + cos(B))
Finally, we can square both sides of the equation:
(tan(B/2)/sin(B))^2 = (1/(1 + cos(B)))^2
tan^2(B/2) / sin^2(B) = 1/(1 + cos(B))^2
Now, using the Pythagorean Identity: sin^2(B) = 1 - cos^2(B), we can replace sin^2(B) in the denominator:
tan^2(B/2) / (1 - cos^2(B)) = 1/(1 + cos(B))^2
Rearranging terms:
tan^2(B/2) = (1 - cos^2(B))/(1 + cos(B))^2
Simplifying the numerator:
tan^2(B/2) = sin^2(B)/(1 + cos(B))^2
And simplifying the denominator:
tan^2(B/2) = sin^2(B)/(1 + 2cos(B) + cos^2(B))
Notice that sin^2(B) / cos^2(B) represents tan^2(B):
tan^2(B/2) = tan^2(B)/(1 + 2cos(B) + cos^2(B))
Therefore, we have proven that (1 - cos(B))/(1 + cos(B)) is equal to tan^2(B/2).