If Tan−1(a3−3aX23a2X−X3)=KTan−1(ax) then k =

tan−1(a3−3ax23a2x−x3)=tan−1(1−3a2x23ax−a3x3)Put ax=tanθ⟹θ=Tan−1axAs we know that tan3θ=1−3tan2θ3tanθ−tan3θ⟹3θ=tan−1(1−3tan2θ3tanθ−tan3θ)⟹tan−1(1−3(ax)23ax−(ax)3)=3tan−1ax⟹Ktan−1ax=3tan−1axK=3

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If Tan^{-1} ( {3a^2X - X^3}{a^3 - 3aX^2}) = K ,, Tan^{-1} ({x}

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If $T a n^{- 1} (\frac{3 a ^{2} X - X ^{3}}{a ^{3} - 3 a X ^{2}}) = K T a n^{- 1} (\frac{x}{a})$ then k =

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T a n^{- 1} (\frac{3 a ^{2} X - X ^{3}}{a ^{3} - 3 a X ^{2}}) = K T a n^{- 1} (\frac{x}{a})

then k =

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Text solutionVerified

tan^{- 1} (\frac{3 a ^{2} x - x ^{3}}{a ^{3} - 3 a x ^{2}}) = tan^{- 1} (\frac{3 \frac{x}{a} - \frac{x ^{3}}{a ^{3}}}{1 - 3 \frac{x ^{2}}{a ^{2}}})

Put

\frac{x}{a} = tan θ ⟹ θ = Tan^{- 1} \frac{x}{a}

As we know that

tan 3 θ = \frac{3 tan θ - tan ^{3} θ}{1 - 3 tan ^{2} θ}

⟹ 3 θ = tan^{- 1} (\frac{3 tan θ - tan ^{3} θ}{1 - 3 tan ^{2} θ})

⟹ tan^{- 1} (\frac{3 \frac{x}{a} - ( \frac{x}{a} ) ^{3}}{1 - 3 ( \frac{x}{a} ) ^{2}}) = 3 tan^{- 1} \frac{x}{a}

⟹ K tan^{- 1} \frac{x}{a} = 3 tan^{- 1} \frac{x}{a}

K = 3

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Question Text	If $T a n^{- 1} (\frac{3 a ^{2} X - X ^{3}}{a ^{3} - 3 a X ^{2}}) = K T a n^{- 1} (\frac{x}{a})$ then k =
Answer Type	Text solution:1
Upvotes	150

If Tan−1(a3−3aX23a2X−X3​)=KTan−1(ax​) then k =

Text solutionVerified

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If $T a n^{- 1} (\frac{3 a ^{2} X - X ^{3}}{a ^{3} - 3 a X ^{2}}) = K T a n^{- 1} (\frac{x}{a})$ then k =