Asked by Lynn
sin^2x/cosx=sec^2x-1
Answers
Answered by
Reiny
sin^2x /cosx = 1/cos^2 x - 1
multiply each side by cos^2 x
sin^2 x cosx = 1 - cos^2x
(1 - cos^2x)cosx = 1 - cos^2x
cosx - cos^3x - 1 + cos^2x = 0
cos^3x - cos^2x - cosx + 1 = 0
cos^2x(cosx - 1) -(cosx - 1) = 0
(cosx-1)(cos^2x-1) =0
(cosx-1)(cosx-1)(cosx+1) = 0
cosx = 1 or cosx = -1
x = 0, 2π or x = π
multiply each side by cos^2 x
sin^2 x cosx = 1 - cos^2x
(1 - cos^2x)cosx = 1 - cos^2x
cosx - cos^3x - 1 + cos^2x = 0
cos^3x - cos^2x - cosx + 1 = 0
cos^2x(cosx - 1) -(cosx - 1) = 0
(cosx-1)(cos^2x-1) =0
(cosx-1)(cosx-1)(cosx+1) = 0
cosx = 1 or cosx = -1
x = 0, 2π or x = π
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