If sin-1x-sin-1y-2- then cos-1x-cos-1y is equal to

Given : sin^ 1x+sin^ 1y=π2⋯ (1) We know that, sin^ 1x+cos^ 1x=π2 ⇒sin^ 1x=π2 cos^ 1x ∴ Equation (1) becomes π2 cos^ 1x+π2 cos^ 1y=π2 ⇒cos^ 1x+cos^ 1y=π2 ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

If sin-1x+sin...

Question

If ${sin}^{- 1} x + {sin}^{- 1} y = \frac{π}{2}$ , then ${cos}^{- 1} x + {cos}^{- 1} y$ is equal to

\frac{3 π}{4}

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π

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\frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is C $\frac{π}{2}$
Given : ${sin}^{- 1} x + {sin}^{- 1} y = \frac{π}{2} \dots (1)$
We know that,
${sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2} \Rightarrow {sin}^{- 1} x = \frac{π}{2} - {cos}^{- 1} x$

$∴$ Equation $(1)$ becomes
$\frac{π}{2} - {cos}^{- 1} x + \frac{π}{2} - {cos}^{- 1} y = \frac{π}{2} \Rightarrow {cos}^{- 1} x + {cos}^{- 1} y = \frac{π}{2}$

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Basic Inverse Trigonometric Functions

Standard XII Mathematics

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If sin−1x+sin−1y=π2, then cos−1x+cos−1y is equal to

The correct option is C π2Given : sin−1x+sin−1y=π2⋯(1) We know that, sin−1x+cos−1x=π2⇒sin−1x=π2−cos−1x ∴ Equation (1) becomes π2−cos−1x+π2−cos−1y=π2⇒cos−1x+cos−1y=π2

If ${sin}^{- 1} x + {sin}^{- 1} y = \frac{π}{2}$ , then ${cos}^{- 1} x + {cos}^{- 1} y$ is equal to