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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`
4 years ago

```							tan(α+β-γ)/tan(α-β+γ)=tanγ/tanβwrite 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 simplifyyou will get …..sin(β-γ)*[2*cos(β-γ)*sin(β+γ)+sin(2α)]=0so 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
```
4 years ago
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