Question 373128: How would you go about doing (1+cosx+sinx)/(1+cosx-sinx)=secx+tanx ?
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! How would you go about doing (1+cosx+sinx)/(1+cosx-sinx)=secx+tanx ?
sohcahtoa
sin = opp/hyp, cos = adj/hyp, tan = opp/adj
sin/cos = opp/hyp * hyp/adj = opp/adj
tan = sin/cos
sec = 1/cos
sin^2(x)=(sinx)^2, cos^2(x) = (cosx)^2
trigonometric identity: (sinx)^2 + (cosx)^2 = 1
(1 + cosx + sinx)/(1 + cosx - sinx) = secx + tanx
put all terms in terms of sin and cos:
(1 + cosx + sinx)/(1 + cosx - sinx) = 1/cosx + sinx/cosx
(1 + cosx + sinx)/(1 + cosx - sinx) = (1 + sinx)/cosx
cross-multiply:
cosx(1 + cosx + sinx) = (1 + sinx)(1 + cosx - sinx)
distribute:
cosx + (cosx)^2 + sinxcosx = 1 + cosx - sinx + sinx + sinxcosx - (sinx)^2
sinx terms on right cancel out
cosx + (cosx)^2 + sinxcosx = 1 + cosx + sinxcosx - (sinx)^2
cosx terms and sinxcosx terms cancel out
(cosx)^2 = 1 - (sinx)^2
(sinx)^2 + (cosx)^2 = 1, done
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