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$I f θ = \tan^{- 1} (2 \tan^{2} θ) - \tan^{- 1} (\frac{1}{3} \tan θ)$ , then $\tan θ =$

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Correct option is A

We have θ=tan−12tan2θ−13tanθ1+23tan3θ ⇒tanθ=6tan2θ−tanθ3+2tan3θ⇒1=6tanθ−13+2tan3θ or tanθ=0⇒2tan3θ−6tanθ+4=0⇒(tanθ−1)2(tanθ+2)=0⇒tanθ=1;tanθ=−2;tanθ=0

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If θ=tan−1(2tan2θ)−tan−1(13tanθ), then tanθ=

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$I f θ = \tan^{- 1} (2 \tan^{2} θ) - \tan^{- 1} (\frac{1}{3} \tan θ)$ , then $\tan θ =$