Asked by Anon
cosx+sinx/cosx-sinx = sec2x+tan2x
Answers
Answered by
mathhelper
I bet you meant:
(cosx+sinx)/(cosx-sinx) = sec(2x) + tan(2x)
and I bet you want to prove that it is an identity, that is
LS = RS
LS = (cosx+sinx)/(cosx-sinx)
= (cosx+sinx)/(cosx-sinx) * (cosx + sinx)/(cosx + sinx)
= (cosx + sinx)^2 / (cos^2 x - sin^2 x)
= (sin^2 x + 2sinxcosx + cos^2 x)/( cos (2x) )
= (1 + sin (2x) )/cos (2x)
= 1/cos(2x) + sin(2x)/cos(2x)
= sec(2x) + sin(2x)
= RS
all done!
(cosx+sinx)/(cosx-sinx) = sec(2x) + tan(2x)
and I bet you want to prove that it is an identity, that is
LS = RS
LS = (cosx+sinx)/(cosx-sinx)
= (cosx+sinx)/(cosx-sinx) * (cosx + sinx)/(cosx + sinx)
= (cosx + sinx)^2 / (cos^2 x - sin^2 x)
= (sin^2 x + 2sinxcosx + cos^2 x)/( cos (2x) )
= (1 + sin (2x) )/cos (2x)
= 1/cos(2x) + sin(2x)/cos(2x)
= sec(2x) + sin(2x)
= RS
all done!
Answered by
Anon
Thank you!
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