Question

. If α + β = θ ; α − β = ∅ and tan α = λ .tanβ . Prove sin θ = ( λ+1 λ−1 ) . sin∅ ?

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \alpha + \beta = \theta$

and

$\rm :\longmapsto\: \alpha - \beta = \phi$

Also, given that

$\rm :\longmapsto\:tan\alpha = \lambda \: tan\beta$

can be rewritten as

$\rm :\longmapsto\:\dfrac{tan\alpha}{tan\beta} = \dfrac{\lambda}{1}$

Apply Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{tan\alpha + tan\beta}{tan\alpha - tan\beta} = \dfrac{\lambda + 1}{\lambda - 1}$

$\rm :\longmapsto\:\dfrac{\dfrac{sin\alpha}{cos\alpha} + \dfrac{sin\beta}{cos\beta} }{\dfrac{sin\alpha}{cos\alpha} - \dfrac{sin\beta}{cos\beta} } = \dfrac{\lambda + 1}{\lambda - 1}$

$\rm :\longmapsto\:\dfrac{sin\alpha \: cos\beta \: + \: sin\beta \: cos\alpha}{sin\alpha \: cos\beta \: - \: sin\beta \: cos\alpha} = \dfrac{\lambda + 1}{\lambda - 1}$

$\rm :\longmapsto\:\dfrac{sin(\alpha + \beta)}{sin(\alpha - \beta)} = \dfrac{\lambda + 1}{\lambda - 1}$

$\rm :\longmapsto\:\dfrac{sin\theta}{sin\phi} = \dfrac{\lambda + 1}{\lambda - 1}$

$\rm :\longmapsto\:sin\theta \: = \: \dfrac{\lambda + 1}{\lambda - 1} \: sin\phi$

Formula Used :-

$\rm :\longmapsto\:sin(x + y) = sincosy + sinycosx$

$\rm :\longmapsto\:sin(x - y) = sincosy - sinycosx$

Additional Information :-

$\rm :\longmapsto\:cos(x + y) = cosxcosy - sinxsiny$

$\rm :\longmapsto\:cos(x - y) = cosxcosy + sinxsiny$

$\rm :\longmapsto\:tan(x + y) = \dfrac{tanx + tany}{1 - tanxtany}$

$\rm :\longmapsto\:tan(x - y) = \dfrac{tanx - tany}{1 + tanxtany}$

$\rm :\longmapsto\:cot(x + y) = \dfrac{cotxcoty - 1}{cotx + coty}$