If sin-1x-sin-1y-sin-1z- then x4-y4-z4-4x2y2z2-kx2y2-y2z2-z2x2- where k is equal to

The correct option is C 2 sin^ 1(x)+sin^ 1(y)=π sin^ 1(z) sin^ 1(x√(1 y^2)+y√(1 x^2))=π sin^ 1(z) x√(1 y^2)+y√(1 x^2)=z x^2(1 y^2)+y^2 ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

If sin-1x+s...

Question

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π$ , then $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ , where $k$ is equal to

1

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Solution

The correct option is C $2$
$s i n^{- 1} (x) + s i n^{- 1} (y) = π - s i n^{- 1} (z)$
$s i n^{- 1} (x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}}) = π - s i n^{- 1} (z)$
$x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}} = z$
$x^{2} (1 - y^{2}) + y^{2} (1 - x^{2}) + 2 x y \sqrt{(1 - x^{2}) (1 - y^{2})} = z^{2}$
$x^{2} + y^{2} - 2 x^{2} y^{2} + 2 x y \sqrt{(1 - x^{2}) (1 - y^{2})} = z^{2}$
$x^{2} + y^{2} - z^{2} = 2 x^{2} y^{2} - 2 x y \sqrt{(1 - x^{2}) (1 - y^{2})}$
Squaring and simplifying, we get
$x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + z^{2} y^{2} + x^{2} z^{2})$

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Similar questions

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π$ then $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ where k is equal to

Q. If

s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}, t h e n \frac{1 + x^{4} + y^{4}}{x^{2} - x^{2} y^{2} + y^{2}}

is equal to

Q. If

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

then the value of

x^{2} + y^{2} + z^{2} + 2 x y z

is equal to

Q. If

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{π}{2}

, then

x^{2} + y^{2} + z^{2} + 2 x y z =

Inverse circular functions,Principal values of

{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

(a) If

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

, then prove that

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

(b) If

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

, then prove that

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

show that

x + y + z = x y z

x y + y z + z x = 1

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If sin−1x+sin−1y+sin−1z=π, then x4+y4+z4+4x2y2z2=k(x2y2+y2z2+z2x2), where k is equal to

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π$ , then $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ , where $k$ is equal to