Solve:
${cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1})$

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Solution

$\begin{matrix} {cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}), l e t x = tan θ, θ = {tan}^{- 1} x \Rightarrow {cos}^{- 1} - (\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ}) = {cos}^{- 1} - (cos 2 θ) [{cos}^{- 1} θ = π - {cos}^{- 1}] \Rightarrow π - {cos}^{- 1} (cos 2 θ) = π - 2 θ = π - {tan}^{- 1} x A n s . \end{matrix}$

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Similar questions

{cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + {tan}^{- 1} \frac{2 x}{x^{2} - 1} = \frac{2 π}{3}

{cos}^{- 1} \frac{x^{2} - 1}{x^{2} + 1} + {tan}^{- 1} \frac{2 x}{x^{2} - 1} = \frac{2 π}{3}

{cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + {sin}^{- 1} (\frac{2 x}{x^{2} + 1}) + {tan}^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3}

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

(where x \in R - {0})

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

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