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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+1k-1 sin Φ

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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

Sum
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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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Chapter 3: Trigonometric Functions - Solved Examples [Page 46]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 3 Trigonometric Functions
Solved Examples | Q 13 | Page 46

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