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Q.

If α,β,γ are acute angles and cos⁡θ=sin⁡β/sin⁡α,cos⁡ϕ=sin⁡γ/sin⁡α and cos⁡(θ−ϕ)=sin⁡βsin⁡γ then the value of tan2⁡α−tan2⁡β−tan2⁡γ 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  cos⁡θcos⁡ϕ+sin⁡θsin⁡ϕ=sin⁡βsin⁡γ⇒ sin2⁡θsin2⁡ϕ=cos⁡θcos⁡ϕ−sin⁡βsin⁡γ)2⇒ 1−sin2⁡βsin2⁡α1−sin2⁡γsin2⁡α=sin⁡βsin⁡γsin2⁡α−sin⁡βsin⁡γ2⇒ sin2⁡α−sin2⁡βsin2⁡α−sin2⁡γ=sin2⁡βsin2⁡γ1−sin2⁡α2⇒ sin4⁡α1−sin2⁡βsin2⁡γ−sin2⁡α sin2⁡β+sin2⁡γ−2sin2⁡βsin2⁡γ=0∴ sin2⁡α=sin2⁡β+sin2⁡γ−2sin2⁡βsin2⁡γ1−sin2⁡βsin2⁡γ and cos2⁡α=1−sin2⁡β−sin2⁡γ+sin2⁡βsin2⁡γ1−sin2⁡βsin2⁡γ⇒ tan2⁡α=sin2⁡β−sin2⁡βsin2⁡γ+sin2⁡γ−sin2⁡βsin2⁡γcos2⁡β−sin2⁡γ1−sin2⁡β=sin2⁡βcos2⁡γ+cos2⁡βsin2⁡γcos2⁡βcos2⁡γ=tan2⁡β+tan2⁡γ⇒ tan2⁡α−tan2⁡β−tan2⁡γ=0
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If α,β,γ are acute angles and cos⁡θ=sin⁡β/sin⁡α,cos⁡ϕ=sin⁡γ/sin⁡α and cos⁡(θ−ϕ)=sin⁡βsin⁡γ then the value of tan2⁡α−tan2⁡β−tan2⁡γ is equal to