Prove that tan−1(1+x2−1−x21+x2+1−x2)=4π+21cos−1x2

To prove:tan−1(1+x2−1−x21+x2+1−x2)=4π+21cos−1x2Proof:Let, x2=cos2θ=(cos2θ−sin2θ)=1−2sin2θ=2cos2θ−1⇒cos−1x2=2θ⇒θ=21cos−1x2∴1+x2=1+cos2θ=1+2cos2θ−1=2cosθAnd 1−x2=1−cos2θ=1−+2sin2θ=2sinθ∴LHS=tan−1(1+x2−1−x21+x2+1−x2)=tan−1(2cosθ−2sinθ2cosθ+2sinθ)=tan−1(cosθ−sinθcosθ+sinθ)=tan−1(1−tanθ1+tanθ)=tan−1⎝⎛1−tan4πtanθtan(4π)+tanθ⎠⎞=tan−1[tan(4π+θ)][∵tan(x+y)=1−tanx.tanytanx+tany]=4π+θ=4π+21cos−1x2=RHS Hence proved.

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Prove that ^{-1}({{1+x^{2}}+{1-x^{2}}}{{1+x^{2}}-{1-x^{2}}})={

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Prove that $tan^{- 1} (\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}}) = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}$

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Prove that

tan^{- 1} (\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}}) = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

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Text solutionVerified

To prove:

tan^{- 1} (\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}}) = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

Proof:

Let,

x^{2} = cos 2 θ = (cos^{2} θ - sin^{2} θ) = 1 - 2 sin^{2} θ = 2 cos^{2} θ - 1

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

∴ 1 + x^{2} = 1 + cos 2 θ

= 1 + 2 cos^{2} θ - 1 = 2 cos θ

And

1 - x^{2} = 1 - cos 2 θ

= 1 - + 2 sin^{2} θ = 2 sin θ

∴ L H S = tan^{- 1} (\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}})

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

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

= tan^{- 1} (\frac{1 + tan θ}{1 - tan θ}) = tan^{- 1} ⎝ ⎛ \frac{tan ( \frac{π}{4} ) + tan θ}{1 - tan \frac{π}{4} tan θ} ⎠ ⎞

= tan^{- 1} [tan (\frac{π}{4} + θ)] [∵ tan (x + y) = \frac{tan x + tan y}{1 - tan x . tan y}]

= \frac{π}{4} + θ = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

= R H S

Hence proved.

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Question Text	Prove that $tan^{- 1} (\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}}) = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}$
Answer Type	Text solution:1
Upvotes	150

Prove that tan−1(1+x2​−1−x2​1+x2​+1−x2​​)=4π​+21​cos−1x2

Text solutionVerified

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Prove that $tan^{- 1} (\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}}) = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}$