An angle theta is divided into two parts alpha and beta such that tan alpha : tan beta = x:y then prove that sin (alpha-beta) =
x-y / x+y sin theta.

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Solution

$tanα : tanβ = x : y \frac{tanα}{tanβ} = \frac{x}{y} . . . (1) Also, θ = α + β . . . . (2) Using componendo - divindo to equation 1, \frac{tanα + tanβ}{tanα - tanβ} = \frac{x + y}{x - y} \frac{x - y}{x + y} = \frac{tanα - tanβ}{tanα + tanβ} = \frac{sinαcosβ - cosαsinβ}{sinαcosβ + cosαsinβ} = \frac{\sin (α - β)}{\sin (α + β)} = \frac{\sin (α - β)}{sinθ} [Using equation 2] So we get \sin (α - β) = (\frac{x - y}{x + y}) sinθ$

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