If an angle - be divided into two parts such that the tangent of one part is - times the tangent of the other part- prove that their difference - is given by sin - - - - 1- - 1sin-

Let α=A+B Given that tan A=λtan B Also, θ=A B Therefore, A=α+θ2; B=α θ2 tan A=λtan B tan (α+θ2)=λtan (α θ2) tan (α+θ2)tan (α ...

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Standard X

Mathematics

Componendo and Dividendo

If an angle ...

Question

If an angle $α$ be divided into two parts such that the tangent of one part is $λ$ times the tangent of the other part, prove that their difference $θ$ is given by sin $θ = \frac{λ - 1}{λ + 1} sin α$ .

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Solution

Let $α = A + B$
Given that $tan A = λ tan B$
Also, $θ = A - B$
Therefore, $A = \frac{α + θ}{2}; B = \frac{α - θ}{2}$
$tan A = λ tan B$
$tan (\frac{α + θ}{2}) = λ tan (\frac{α - θ}{2})$
$\frac{tan (\frac{α + θ}{2})}{tan (\frac{α - θ}{2})} = λ$
Converting into $sin, cos$ and by applying componendo-dividendo, we get
$sin θ = \frac{λ - 1}{λ + 1} sin (α)$

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If an angle α be divided into two parts such that the tangent of one part is λ times the tangent of the other part, prove that their difference θ is given by sin θ=λ−1λ+1sinα.

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If an angle $α$ be divided into two parts such that the tangent of one part is $λ$ times the tangent of the other part, prove that their difference $θ$ is given by sin $θ = \frac{λ - 1}{λ + 1} sin α$ .