If θ1 and θ2 are two values lying in [0,2π] for which tanθ=λ, then tanθ12tanθ22 is equal to
0
-1
2
1
tanθ=λ we get 2tanθ/21−tan2θ/2=λ
or λtan2θ2+2tanθ2−λ=0⇒ tanθ12tanθ22=−1