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Mathematics 10 Online
OpenStudy (anonymous):

how do you verify csc^4x - 2csc^2x + 1 = cot^4x?

OpenStudy (ajprincess):

1-sin^2x=cos^2x cotx=cosx/sinx First take LHS csc^4x - 2csc^2x + 1 =(csc^2x-1)^2 =(1/sin^2x-1)^2 =((1-sin^2x)/sin^2x)^2 =(cos^2x/sin^2x)^2 =(cot^2x)^2 =cot^4x

OpenStudy (ajprincess):

LHS=RHS Hence the result

OpenStudy (ajprincess):

Is that clear @jamroz ?

OpenStudy (anonymous):

i don't understand how you got (csc^2x-1)^2? @ajprincess

OpenStudy (ajprincess):

csc^4x-2csc^2x+1 =csc^4x-csc^2x-csc^2x+1 =csc^2x(csc^2x-1)-1(csc^2x-1) =(csc^2x-1)(csc^2x-1) =(csc^2x-1)^2 Is that clear nw?

OpenStudy (ajprincess):

@jamroz is that clear nw?

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