${sin}^{- 1} (cos ({sin}^{- 1} x)) + {cos}^{- 1} (sin ({cos}^{- 1} x)) is equal to .$

\frac{π}{2}

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\frac{π}{4}

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\frac{3 π}{4}

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Solution

The correct option is B $\frac{π}{2}$
Given

${sin}^{- 1} (cos {sin}^{- 1} (x)) + {cos}^{- 1} (sin ({cos}^{- 1} x))$

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

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

$π - {sin}^{- 1} x - {cos}^{- 1} x$

$({cos}^{- 1} cos x = x, {sin}^{- 1} sin x = x)$

$π - ({sin}^{- 1} x + {cos}^{- 1} x)$

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

$⟹ \frac{π}{2}$

Suggest Corrections

Similar questions

α = {sin}^{- 1} (cos ({sin}^{- 1} x))

β = {cos}^{- 1} (sin ({cos}^{- 1} x))

{sin}^{- 1} cos ({sin}^{- 1} x) + {cos}^{- 1} sin ({cos}^{- 1} x)

| x | \leq 1

α = {sin}^{- 1} (cos ({sin}^{- 1} x))

β = {cos}^{- 1} (sin ({cos}^{- 1} x))

tan α tan β

{sin}^{- 1} x \in (0, \frac{π}{2})

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

{cos}^{- 1} (sin ({cos}^{- 1} x)) = \frac{π}{6}

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