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sin(a+b)/sin(a-b) = tan(a)+tan(b)/tan(a)-tan(b)
In order to prove the trigonometric identity sin(a+b)/sin(a-b) = tan(a)+tan(b)/tan(a)-tan(b), we need to use the trigonometric identities for sin and tan.
Let's start by expressing sin(a+b) and sin(a-b) in terms of sine and cosine functions:
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
Now, let's substitute these expressions into the left side of the equation:
sin(a+b)/sin(a-b) = (sin(a)cos(b) + cos(a)sin(b)) / (sin(a)cos(b) - cos(a)sin(b))
Next, let's simplify this expression by multiplying both the numerator and denominator by the conjugate of the denominator:
sin(a+b)/sin(a-b) = [(sin(a)cos(b) + cos(a)sin(b))(sin(a)cos(b) + cos(a)sin(b))] / [(sin(a)cos(b) - cos(a)sin(b))(sin(a)cos(b) + cos(a)sin(b))]
sin(a+b)/sin(a-b) = (sin^2(a)cos^2(b) + 2sin(a)cos(a)sin(b)cos(b) + cos^2(a)sin^2(b)) / (sin^2(a)cos^2(b) - cos^2(a)sin^2(b))
Now, let's use the trigonometric identity tan(x) = sin(x)/cos(x) to express tan(a) and tan(b):
tan(a) = sin(a)/cos(a)
tan(b) = sin(b)/cos(b)
Substitute these expressions into the right side of the equation:
(tan(a) + tan(b))/(tan(a) - tan(b)) = (sin(a)/cos(a) + sin(b)/cos(b)) / (sin(a)/cos(a) - sin(b)/cos(b))
(tan(a) + tan(b))/(tan(a) - tan(b)) = [(sin(a)cos(b) + cos(a)sin(b)) / (cos(a)cos(b))] / [(sin(a)cos(b) - cos(a)sin(b)) / (cos(a)cos(b))]
(tan(a) + tan(b))/(tan(a) - tan(b)) = (sin(a)cos(b) + cos(a)sin(b)) / (sin(a)cos(b) - cos(a)sin(b))
Now, we can see that the left side of the equation is equal to the right side of the equation, proving the trigonometric identity sin(a+b)/sin(a-b) = tan(a)+tan(b)/tan(a)-tan(b).