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If tan(α+β-γ)/tan(α-β+γ)=tanγ/tanβ then prove that sin(β-γ)=0 or sin2α+sin2β+sin2γ=0. Please explain with full steps

If tan(α+β-γ)/tan(α-β+γ)=tanγ/tanβ then prove that sin(β-γ)=0 or sin2α+sin2β+sin2γ=0.

Please explain with full steps

Grade:11

1 Answers

Akshay
185 Points
8 years ago
tan(α+β-γ)/tan(α-β+γ)=tanγ/tanβ
write in terms of sine and cos,
sin(α+β-γ) * cos(α-β+γ) / (cos(α+β-γ) * sin(α-β+γ)) = sinγ * cosβ / (cosγ * sinβ),   (1)
write sin(α+β-γ) * cos(α-(β-γ)) = ½ * [sin(2α) + sin(2(β-γ)) ] ,
and write  cos(α+β-γ) * sin(α-β+γ) = ½ * [sin(2α) - sin(2(β-γ)) ] ,
now apply ratio property that if A/B=C/D, then (A-B)/(A+B) = (C-D)/(C+D),
(1) will become, sin(2(β-γ)) / sin(2α) = – sin(β-γ)/sin(β+γ), [using sin(x+y) and sin(x-y) properties in LHS],
now, write sin(2(β-γ)) = 2*sin(β-γ)*cos(β-γ), and simplify
you will get …..sin(β-γ)*[2*cos(β-γ)*sin(β+γ)+sin(2α)]=0
so either sin(β-γ)=0 or [2*cos(β-γ)*sin(β+γ)+sin(2α)]=0,
now [2*cos(β-γ)*sin(β+γ)]=sin(2β)+sin(2γ)...use sinx+siny property,
second solution will become sin(2β)+sin(2γ)+sin(2α)=0
 

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