If `(sin^(-1)x+sin^(-1)w)(sin^(-1)y+sin^(-1)z)=pi^2,` then `D=|x^(N_1)y^(N_3)z^(N_3)w^(N_4)|(N_1,N_2,N_3,N_4 in N)` has a maximum value of 2 has a maximum value of 0 16 different D are possible has a minimum value of `-2`

Explanation

`(sin^-1x+sin^-1w)(sin^-1y+sin^-1z) = pi^2`

Maximum value of `sin^-1x` is `pi/2` and minimum value of `sin^-1x` is `-pi/2`.

So, the solution for the given equation can be,

`x = y = z = w = 1` or `x = y = z = w = -1`

Now, `D = |[x^(N_1),y^(N_2)],[z^(N_3),w^(N_4)]|`

Maximum value of `D` will be when `y^(N_2) = -1`

`:. D = |[1,-1],[1,1]| = 2`

Minimum value of `D` will be when `x^(N_1) = -1`

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