${sin}^{- 1} (sin 5) + {cos}^{- 1} (cos 7) - {tan}^{- 1} (tan 5) =$

7 - 2 π

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π

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3 π - 1

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2 π - 1

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Solution

The correct option is D $7 - 2 π$
As ${sin}^{- 1} (sin θ) = θ$ for $θ \in [- \frac{π}{2}, \frac{π}{2}];$
${cos}^{- 1} (cos θ) = θ$ for $θ \in [0, π]$
and ${tan}^{- 1} (tan θ) = θ$ for $θ \in (- \frac{π}{2}, \frac{π}{2})$
$∴ {sin}^{- 1} (sin 5) + {cos}^{- 1} (cos 7) - {tan}^{- 1} (tan 5)$
$= {sin}^{- 1} (sin (5 - 2 π)) + {cos}^{- 1} (cos (7 - 2 π)) - {tan}^{- 1} (tan (5 - 2 π))$
$= 5 - 2 π + 7 - 2 π - 5 + 2 π$
$= 7 - 2 π$

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