If sin^{–1}x + sin^{–1}y + sin^{–1}z + sin^{–1}t = 2π, then find the value of

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

Range of sin^{–1}x is .

Give that sin^{–1}x + sin^{–1}y + sin^{–1}z + sin^{–1}t = 2π

Each of sin^{–1}x, sin^{–1}y, sin^{–1}z, sin^{–1}t takes value of .

So,

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

Hence,

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

= 1 + 1 + 1 + 1

= 4

4