If sin-1x + sin-1y+ sin-1z=π , then x4+y4+z4+4x2 y2 z2=K(x2 y2+y2 z2+z2 x2) , where K =

Question

If sin-1x + sin-1y+ sin-1z=π , then x4+y4+z4+4x2 y2 z2=K(x2 y2+y2 z2+z2 x2) , where K = (A) 1 (B) 2 (C) 4 (D) none of these.

Tardigrade Admin · Accepted Answer

Since sin-1x+sin-1 y = π-sin-1 z = π ∴ sin-1 x +sin-1 y = π - sin-1z ⇒ sin-1[x√1-y2 +y√1-x2] = π -sin-1(z) ⇒ x √1-y2 + y √1-x2 = sin (π-sin-1 (z)) = sin (sin-1 z) = z ⇒ x2 (1-y2) = z2 +y2(1-x2)-2zy √1-x2 ⇒ (x2+y2-z2) = 4y2z2(1-x2) ⇒ x4+y4+z4-2x2y2 -2x2z2 +2y2z2 = 4y2z2-4x2y2z2 ⇒ x4+y4+z4 +4x2y2z2 = 2(x2y2+y2z2+z2x2) ∴ K = 2

Q. If sin−1x+sin−1y+sin−1z=π , then x4+y4+z4+4x2y2z2=K(x2y2+y2z2+z2x2) , where K =

1

2

4

none of these.

Q. If $sin^{- 1} x + sin^{- 1} y + sin^{- 1} z = π$ , 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 =