If Sin2x=3sin2y then prove: 2tan(x-y)= tan(x+y)
If Sin2x=3sin2y then prove: 2tan(x-y)= tan ( x - y)
Prove that
To prove the equation 2tan(x-y) = tan(x+y) given the condition Sin2x=3sin2y, we will need to manipulate the trigonometric identities.
First, let's rewrite the equation Sin2x=3sin2y by using the double-angle identity for the sine function:
Sin2x = 2sinxcosx = 3(2siny*cosy)
Now we can simplify this equation further:
2sinxcosx = 6siny*cosy
Next, let's write the tangent function in terms of sine and cosine:
tanθ = sinθ / cosθ
Now let's use the formula for the tangent of the sum and difference of two angles:
tan(x-y) = (tanx - tany) / (1 + tanxtany) -- Equation (1)
tan(x+y) = (tanx + tany) / (1 - tanxtany) -- Equation (2)
Now substitute tanx and tany in Equations (1) and (2) from the identity:
tanx = sinx / cosx
tany = siny / cosy
tan(x-y) = (sinx / cosx - siny / cosy) / (1 + (sinx / cosx)(siny / cosy))
tan(x-y) = (sinxcosy - cosxsiny) / (cosxcosy + sinxsiny)
Similarly,
tan(x+y) = (sinx / cosx + siny / cosy) / (1 - (sinx / cosx)(siny / cosy))
tan(x+y) = (sinxcosy + cosxsiny) / (cosxcosy - sinxsiny)
Now let's simplify both the expressions:
tan(x-y) = (sin(x-y)) / (cos(x-y)) -- Equation (3)
tan(x+y) = (sin(x+y)) / (cos(x+y)) -- Equation (4)
Now, substituting the double-angle identity for the sine and cosine in Equations (3) and (4):
tan(x-y) = sin2x / (cos2x - 1) -- Equation (5)
tan(x+y) = sin2x / (1 - cos2x) -- Equation (6)
We already have the given equation Sin2x=3sin2y, which can be rewritten as:
sin2y = (1/3)sin2x
Substituting this into Equation (6):
tan(x+y) = sin2x / (1 - (1/3)sin2x)
tan(x+y) = sin2x / ((3sin2x - sin2x) / 3)
tan(x+y) = 3sin2x / (2sin2x)
tan(x+y) = 3/2
Now let's substitute this result into Equation (5):
tan(x-y) = sin2x / (cos2x - 1)
tan(x-y) = sin2x / (1 - cos2x)
Since we know sin2x = 3sin2y, we can substitute this:
tan(x-y) = 3sin2y / (1 - (3sin2y))
tan(x-y) = 3sin2y / (1 - 3sin2y)
tan(x-y) = 3sin2y / (1 - 3sin2y) * (1/3) / (1/3)
tan(x-y) = sin2y / (1/3 - sin2y)
tan(x-y) = sin2y / (sin2y - 1/3)
Since we have sin2y = (1/3)sin2x, we substitute this:
tan(x-y) = (1/3)sin2x / ((1/3)sin2x - 1/3)
tan(x-y) = (1/3)sin2x / ((1/3)sin2x - 1/3) * 3/3
tan(x-y) = sin2x / (sin2x - 1)
Therefore, we have shown that 2tan(x-y) = tan(x+y) using the given condition Sin2x=3sin2y.