Asked by Anonymous
(sinx-cosx+1)/(sinx+cosx-1)=(sinx+1)/(cosx)
Answers
Answered by
Steve
you have
(sinx-(cosx-1))/(sinx+(cosx-1))
multiply top and bottom by
(sinx-(cosx-1)) and you have
(sinx-(cosx-1))^2/(sin^2x - (cosx-1)^2)
(sin^2x - 2sinx(cosx-1) + (cosx-1)^2)
---------------------------------------
(sin^2x - cos^2x + 2cosx - 1)
(sin^2x - 2sinx*cosx + 2sinx + cos^2x - 2cosx + 1)
----------------------------------------
(sin^2x - cos^2x + 2cosx - 1)
2(1 - sinx*cosx + sinx - cosx)
---------------------------------
2(cosx - cos^2x)
((1+sinx) - cosx(1+sinx))
------------------------------
cosx(1-cosx)
(1+sinx)(1-cosx)
-------------------
cosx(1-cosx)
(1+sinx)/cosx
(sinx-(cosx-1))/(sinx+(cosx-1))
multiply top and bottom by
(sinx-(cosx-1)) and you have
(sinx-(cosx-1))^2/(sin^2x - (cosx-1)^2)
(sin^2x - 2sinx(cosx-1) + (cosx-1)^2)
---------------------------------------
(sin^2x - cos^2x + 2cosx - 1)
(sin^2x - 2sinx*cosx + 2sinx + cos^2x - 2cosx + 1)
----------------------------------------
(sin^2x - cos^2x + 2cosx - 1)
2(1 - sinx*cosx + sinx - cosx)
---------------------------------
2(cosx - cos^2x)
((1+sinx) - cosx(1+sinx))
------------------------------
cosx(1-cosx)
(1+sinx)(1-cosx)
-------------------
cosx(1-cosx)
(1+sinx)/cosx
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