Asked by urgent
prove that
(cosx - sinx) / (cosx + sinx) = sec2x - tan2x
(cosx - sinx) / (cosx + sinx) = sec2x - tan2x
Answers
Answered by
oobleck
multiply top and bottom by (cosx - sinx) to get
(cosx-sinx)^2 / (cos^2x - sin^2x)
(cos^2x - 2sinx cosx + sin^2x) / cos2x
(1 - sin2x)/cos2x
1/cos2x - sin2x/cos2x
QED
(cosx-sinx)^2 / (cos^2x - sin^2x)
(cos^2x - 2sinx cosx + sin^2x) / cos2x
(1 - sin2x)/cos2x
1/cos2x - sin2x/cos2x
QED
Answered by
Anonymous
(cosx - sinx)^2 / (cosx + sinx)(cosx - sin x)
= (cos^2x - 2 sinx cosx + sin^2x) / (cos^2x - sin^2 x)
= (1 -2 sinx cosx) / cos2x
= 1/cos2x - sin2x /cos2x
= sec2x - tan2x
whew !
= (cos^2x - 2 sinx cosx + sin^2x) / (cos^2x - sin^2 x)
= (1 -2 sinx cosx) / cos2x
= 1/cos2x - sin2x /cos2x
= sec2x - tan2x
whew !
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