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|[1,1,1],[1,1+x,1],[1,1,1+y]|=xy...

|[1,1,1],[1,1+x,1],[1,1,1+y]|=xy

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|(1+x,1,1),(1,1+y,1),(1,1,1+z)|=xy+yz+zx+xyz

|{:(1+x,1,1),(1,1+y,1),(1,1,1+z):}|=xy+yz+zx+xyz

1+x,1,11,1+y,11,1,1+z]|=xy+yz+zx+xyz

Value of the determinant |(x,1,1),(0,1+x, 1),(-y, 1+x, 1+y)| is (A) xy (B) xy(x+2) (C) x(x+1)(y+1) (D) xy(x+1)

Value of the determinant |(x,1,1),(0,1+x, 1),(-y, 1+x, 1+y)| is (A) xy (B) xy(x+2) (C) x(x+1)(y+1) (D) xy(x+1)

solve |[1,yz,yz(y+z)],[1,zx,zx(z+x)],[1,xy,xy(x+y)]|

If D_1=|{:(1,yz,x),(1,zx,y),(1,xy,z):}|and D_2=|{:(1,1,1),(x,y,z),(x^2,y^2,z^2):}| then ,………

If A(x)=(1)/(sqrt(1-x^(2)))[(1,-x),(-x,1)] , prove that : A(x)A(y)=A((x+y),(1+xy)) , where |x|lt1 . Hence, deduce that (A(x))^(-1)=A(-x) .

If tan^(-1) x +tan^(-1) y = pi/4 , xy lt 1 , then prove that x+y+xy=1 .