If 0<x<1 , then tan−1(1+x1−x2) is equal to

tan−1(1+x1−x2)Let x=cosθ=tan−1(1+cosθ1−cos2θ)tan−11+cos22θ−sin22θsinθ=tan−12cos22θ2sin2θcos2θ=tan−1(tan2θ)=2θ=21cos−1x

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If 0< x < 1 , then ^{-1}({{1-x^{2}}}{1+x}) is equal to

Question

If $0 < x < 1$ , then $tan^{- 1} (\frac{1 - x ^{2}}{1 + x})$ is equal to

\frac{1}{2} cos^{- 1} x

cos^{- 1} \frac{1 + x}{2}

sin^{- 1} \frac{1 - x}{2}

\frac{1}{2} \frac{1 + x}{1 - x}

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Amelia Discussed

0 < x < 1

, then

tan^{- 1} (\frac{1 - x ^{2}}{1 + x})

is equal to

9 mins ago

Discuss this question LIVE

9 mins ago

tan^{- 1} (\frac{1 - x ^{2}}{1 + x})

Let

x = cos θ

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

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

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

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

\frac{1}{2} cos^{- 1} x

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Question Text	If $0 < x < 1$ , then $tan^{- 1} (\frac{1 - x ^{2}}{1 + x})$ is equal to
Answer Type	Text solution:1
Upvotes	150