Solve tan−1{1+x2+1−x21+x2−1−x2}

Let x2=cos2θ ⇒2θ =cos−1x2⇒ θ =21cos−1x2=tan1{1+cos2θ+1−cos2θ1+cos2θ−1−cos2θ}=tan−1{23cosθ +2sinθ2cosθ−2sinθ} divided each them by 2cosθ=tan−1{1+tanθ1−tanθ}=tan−1tan(4π−θ)

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Solve ^{-1} { {{1 +x^{2} - {1 - x^{2}}}}{{1 + x^{2} + {1 -

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Solve $tan^{- 1} {\frac{1 + x ^{2} - 1 - x ^{2}}{1 + x ^{2} + 1 - x ^{2}}}$

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Text solutionVerified

Let

x^{2} = cos 2 θ

\Rightarrow 2 θ = cos^{- 1} x^{2} \Rightarrow θ = \frac{1}{2} cos^{- 1} x^{2}

= tan^{1} {\frac{1 + cos 2 θ - 1 - cos 2 θ}{1 + cos 2 θ + 1 - cos 2 θ}}

= tan^{- 1} {\frac{2 cos θ - 2 sin θ}{23 cos θ + 2 sin θ}}

divided each them by

2 cos θ

= tan^{- 1} {\frac{1 - tan θ}{1 + tan θ}}

= tan^{- 1} tan (\frac{π}{4} - θ)

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Question Text	Solve $tan^{- 1} {\frac{1 + x ^{2} - 1 - x ^{2}}{1 + x ^{2} + 1 - x ^{2}}}$
Answer Type	Text solution:1
Upvotes	150

Solve tan−1{1+x2+1−x2​​1+x2−1−x2​​​}

Text solutionVerified

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Solve $tan^{- 1} {\frac{1 + x ^{2} - 1 - x ^{2}}{1 + x ^{2} + 1 - x ^{2}}}$