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प्रश्न
If `sin^-1 x + sin^-1 y+sin^-1 z+sin^-1 t=2pi` , then find the value of x2 + y2 + z2 + t2
उत्तर
We know that the maximum value of sin-1 x, sin-1 y, sin-1 z and sin-1 t is `pi/2`
Now,
LHS = `sin^-1 x+sin^-1y+sin^-1z+sin^-1t`
`=pi/2+pi/2+pi/2+pi/2`
= 2 π = RHS
Now,
`sin^-1x=pi/2,sin^-1y=pi/2,sin^-1z=pi/2` and `sin^-1t=pi/2`
⇒ `x = sin x/2,y=sin pi/2, z=sin pi/2 and t=sin pi/2`
⇒ x = 1, y = 1, z = 1, and t = 1
∴ x2 + y2 + z2 + t2 = 1 + 1 + 1 + 1 = 4
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