If sin-1 x + sin-1 y + sin-1z = (π/2) then value of x2 + y2 + z2 + 2xyz equals

Question

If sin-1 x + sin-1 y + sin-1z = (π/2) then value of x2 + y2 + z2 + 2xyz equals (A) 2 (B) 0 (C) -1 (D) 1

Tardigrade Admin · Accepted Answer

sin-1 x + sin-1 z = (π/2) - sin-1y Using cos on both sides, we have √1 - x2 √1 - z2 = xz + y ⇒ x2 + y2 + z2 + 2xyz = 1 (squaring and adjusting the terms) Short Cut Method: sin-1 x + sin-1y + sin-1 z = (π/2) ⇒ sin-1 x = sin-1y = sin-1 z = (π/2) × (1/3) x = y = z = (1/2) ∴ x2 + y2 + z2 + 2xyz = (3/4) + (1/4) = 1

Q. If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{π}{2}$ then value of $x^{2} + y^{2} + z^{2} + 2 x yz$ equals

$2$

$0$

$- 1$

$1$

Q. If sin−1x+sin−1y+sin−1z=2π​ then value of x2+y2+z2+2xyz equals

2

0

−1

1

sin−1x+sin−1z=2π​−sin−1y Using cos on both sides, we have 1−x2​1−z2​=xz+y ⇒x2+y2+z2+2xyz=1 (squaring and adjusting the terms) Short Cut Method : sin−1x+sin−1y+sin−1z=2π​ ⇒sin−1x=sin−1y=sin−1z=2π​×31​ x=y=z=21​ ∴x2+y2+z2+2xyz =43​+41​=1

Q. If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{π}{2}$ then value of $x^{2} + y^{2} + z^{2} + 2 x yz$ equals

$2$

$0$

$- 1$

$1$