Prove that:cos−1x+x−1x−x−1=2tan−1x1

LHS=cos−1(x+x−1x−x−1)=cos−1(1+x211−x21)Put x1=tanθ⟹θ=tan−1x1⟹LHS=cos−1(1+tan2θ1−tan2θ)=cos−1(cos2θ)=2θ=2tan−1(x1)

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Prove that: ^{-1} {x-x^{-1}}{x+x^{-1}}=2^{-1}{1}{x}

Question

Prove that:
$cos^{- 1} \frac{x - x ^{- 1}}{x + x ^{- 1}} = 2 tan^{- 1} \frac{1}{x}$

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Prove that:

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

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

Put

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

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

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