For $$0$$<θ<π$$2$$,$$tan$$θ$$+tan2$$θ$$+tan3$$θ$$=0$$ if

Correct option is 4tanθ.tan2θ=2

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For $0 < θ < \frac{π}{2}, \tan θ + \tan 2 θ + \tan 3 θ = 0$ if

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tanθ=0

tan2θ=0

tan3θ=2

tanθ.tan2θ=2

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detailed solution

Correct option is D

Clearly, tanθ≠0 and tan2θ≠0 for 0<θ<π2We havetan3θ=tan2θ+θ⇒tan3θ=tan2θ+tanθ1−tan2θtanθNow 0=tanθ+tan2θ+tan3θ = tan3θ1−tan2θtanθ+tan3θ = tan3θ2−tan2θtanθ⇒tan3θ=0 (or) tan2θtanθ=2

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For 0<θ<π2,tanθ+tan2θ+tan3θ=0 if

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For $0 < θ < \frac{π}{2}, \tan θ + \tan 2 θ + \tan 3 θ = 0$ if