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Question
If angle θ is divided into two parts such that the tangent of one part is k times the tangent of other, and Φ is their difference, then show that sin θ = `(k + 1)/(k - 1)` sin Φ
Solution
Let θ = α + β.
Then tan α = k tan β
or `tanalpha/tanbeta = k/1`
Applying componendo and dividendo, we have
`(tan alpha + tan beta)/(tan alpha - tan beta) = (k + 1)/(k - 1)`
or `(sin alpha cos beta + cos alpha sin beta)/(sin alpha cos beta - cos alpha sin beta) = (k + 1)/(k - 1)`
i.e., `(sin(alpha + beta))/(sin(alpha - beta)) = (k + 1)/(k - 1)`
Given that α – β = Φ and α + β = θ.
Therefore, `sin θ/sin phi = (k + 1)/(k - 1)`
or sin θ = `(k + 1)/(k - 1) sin phi`
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