If sin–1x + sin–1y + sin–1z + sin–1t = 2π, then find the value of
x2 + y2 + z2 + t2.
Range of sin–1x is 
.
Give that sin–1x + sin–1y + sin–1z + sin–1t = 2π
Each of sin–1x, sin–1y, sin–1z, sin–1t takes value of 
.
So,
x = 1, y = 1, z = 1 and t = 1.
Hence,
= x2 + y2 + z2 + t2
= 1 + 1 + 1 + 1
= 4
4