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Q.

 If cosθ1θ2cosθ1+θ2+cosθ3+θ4cosθ3θ4=0 then tanθ1tanθ2tanθ3tanθ4=

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a

2

b

1

c

-1

d

0

answer is C.

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Detailed Solution

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cosθ1θ2cosθ1+θ2+cosθ3+θ4cosθ3θ4=0 cos(θ1θ2)cos(θ1+θ2)=-cos(θ3+θ4)cos(θ3θ4) using componendo and dividendo cos(θ1θ2)+cos(θ1+θ2)cos(θ1θ2)-cos(θ1+θ2)=-cos(θ3+θ4)+cos(θ3θ4)-cos(θ3+θ4)-cos(θ3θ4) 2cosθ1cosθ22sinθ1sinθ2=2sinθ3sinθ4-2cosθ3cosθ4 1tanθ1tanθ2=-tanθ3tanθ4 tanθ1tanθ2tanθ3tanθ4=-1 

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