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If `x+y+z=pi/2,`then prove that`|[sinx,siny,sinz],[cosx,cosy,cosz],[cos^3x,cos^3 y,cos^3z]|=0`

Answer» `L.H.S. = |[sinx,siny,sinz],[cosx,cosy,cosz],[cos^3x,cos^3y,cos^3z]|`
By expending the determinant along `R_3`,
`=sum cos^3x[sinycosz - cosysinz]`
`=sum cos^3xsin(y-z)`
As, `x+y+z = pi/2`, so the determinant becomes,
`=sum cos^3(pi/2-(y+z))sin(y-z)`
`=sum sin^3(y+z)sin(y-z)`
`=sum sin^2(y+z) sin(y+z)sin(y-z)`
`=sum sin^2(pi/2-x) (sin^2y-sin^2z)`
`=sum cos^2x (sin^2y-sin^2z)`
`=sum (1-sin^2x)(sin^2y-sin^2z)`
Let `sin^2x = a, sin^2y = b, sin^2z = c`, then, our determinant becomes,
`=sum(1-a)(b-c)`
`=(1-a)(b-c)+(1-b)(c-a)+(1-c)(a-b)`
`=(b-c+c-a+a-b)-ab+ac-bc+ab-ac+bc`
`=0 = R.H.S.`


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