If sin -1 x-sin -1 y-sin -1 z- then x4-y4-z4-4 x2 y2 z2-kx2 y2-y2 z2-z2 x2 where k is equal to

The correct option is A 2sin^ 1x+sin^ 1y=π sin^ 1z Or x√(1 y^2)+y√(1 x^2)=z Or x^2(1 y^2)=2^2+y^2(1 x^2) 2yz√(1 x^2) Or (x^2 z^2 y^2)^2=4y^2z^2(1 x^2) Or x^4+y ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If sin -1 x+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

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Solution

The correct option is B
2

$s i n^{- 1} x + s i n^{- 1} y = π - s i n^{- 1} z$
Or $x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}} = z$
Or $x^{2} (1 - y^{2}) = 2^{2} + y^{2} (1 - x^{2}) - 2 y z \sqrt{1 - x^{2}}$
Or $(x^{2} - z^{2} - y^{2})^{2} = 4 y^{2} z^{2} (1 - x^{2})$
Or $x^{4} + y^{4} + z^{4} - 2 x^{2} z^{2} + 2 y^{2} z^{2} - 2 x^{2} y^{2} + 4 x^{2} y^{2} z^{2} - 4 y^{2} z^{2} = 0$
Or $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = \frac{2}{x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}}$
$∴ k = 2$

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Similar questions

Q. 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

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

Q. If

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

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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

The correct option is B 2sin−1x+sin−1y=π−sin−1z Or x√1−y2+y√1−x2=z Or x2(1−y2)=22+y2(1−x2)−2yz√1−x2 Or (x2−z2−y2)2=4y2z2(1−x2) Or x4+y4+z4−2x2z2+2y2z2−2x2y2+4x2y2z2−4y2z2=0 Or x4+y4+z4+4x2y2z2=2x2y2+y2z2+z2x2 ∴k=2

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