If α,β,γ are acute angles and cosθ=sinβ/sinα,cosϕ=sinγ/sinα and cos(θ−ϕ)=sinβsinγ then the value of tan2α−tan2β−tan2γ is equal to
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a
-1
b
0
c
1
d
2
answer is B.
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Detailed Solution
From the third relation we get cosθcosϕ+sinθsinϕ=sinβsinγ⇒ sin2θsin2ϕ=cosθcosϕ−sinβsinγ)2⇒ 1−sin2βsin2α1−sin2γsin2α=sinβsinγsin2α−sinβsinγ2⇒ sin2α−sin2βsin2α−sin2γ=sin2βsin2γ1−sin2α2⇒ sin4α1−sin2βsin2γ−sin2α sin2β+sin2γ−2sin2βsin2γ=0∴ sin2α=sin2β+sin2γ−2sin2βsin2γ1−sin2βsin2γ and cos2α=1−sin2β−sin2γ+sin2βsin2γ1−sin2βsin2γ⇒ tan2α=sin2β−sin2βsin2γ+sin2γ−sin2βsin2γcos2β−sin2γ1−sin2β=sin2βcos2γ+cos2βsin2γcos2βcos2γ=tan2β+tan2γ⇒ tan2α−tan2β−tan2γ=0