If (sin–1x)2 + (sin–1y)2 + (sin–1z)2 = 3/4 π2. Find x2 + y2 +

1.	If (sin–1x)2 + (sin–1y)2 + (sin–1z)2 = 3/4 π2. Find x2 + y2 + z2.
Answer» Range of sin^-1x is \([-\frac{\pi}2,\frac{\pi}2].\) Given that (sin^-1x)² + (sin^-1y)² + (sin^-1z)² = \(\frac 34\)π² Each of sin^-1x, sin^-1y and sin^-1z takes the value of \(\frac{\pi}2.\) x = 1, y = 1, and z = 1. Hence, = x² + y² + z² = 1 + 1 + 1 = 3.

If (sin–1x)2 + (sin–1y)2 + (sin–1z)2 = 3/4 π2. Find x2 + y2 + z2.

Answer»

Range of sin^-1x is \([-\frac{\pi}2,\frac{\pi}2].\)

Given that

(sin^-1x)² + (sin^-1y)² + (sin^-1z)² = \(\frac 34\)π²

Each of sin^-1x, sin^-1y and sin^-1z takes the value of \(\frac{\pi}2.\)

x = 1, y = 1, and z = 1.

Hence,

= x² + y² + z²

= 1 + 1 + 1

= 3.