If sin -1 x -sin -1 y -2- then cos -1 x -cos -1 y is equal toA- - - 2B- - - 4C- 3 - - 4D- -

The correct option is A π/2sin^ 1x+sin^ 1y=π/2 ∴π/2 cos^ 1x+π/2 cos^ 1y=π/2 ⇒ cos^ 1x+cos^ 1y=π/2 ...

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Parametric Differentiation

If sin -1 x +...

Question

If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ ,then $c o s^{- 1} x + c o s^{- 1} y$ is equal to

π/2

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π/4

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3π/4

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Similar questions

{sin}^{- 1} x + {sin}^{- 1} y = \frac{π}{2}

{cos}^{- 1} x + {cos}^{- 1} y

c o s^{- 1} x + c o s^{- 1} y = \frac{π}{2}

c o s^{- 1} x = s i n^{- 1} y

{sin}^{- 1} x - {sin}^{- 1} y = \frac{2 π}{3}; {cos}^{- 1} x - {cos}^{- 1} = \frac{π}{3}

x

y

{sin}^{- 1} x + {sin}^{- 1} y = \frac{2 π}{3}

{cos}^{- 1} x - {cos}^{- 1} y = \frac{π}{3}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

p^{2} + q^{2} + r^{2} + 2 p q r = 1

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

{t a n}^{- 1} x + {t a n}^{- 1} y + {t a n}^{- 1} z = π

π / 2

x + y + z = x y z

x y + y z + z x = 1

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