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Verify identities in trigo: (sin x- cos x+1)/(sin x + cos x -1) = (sin x +1)/cos x
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another way: rewrite numerator as ((sinx+1)-cosx)...now multiply both top and bottom by ((sinx+1)+cosx)..the numerator will be the difference between two squares. if you mulitply out u shud get ((sinx)^2 + 2sinx +1-(cosx)^2))/ 2sinxcosx....now we know (sinx)^2+(cosx)^2 = 1, so replace 1 in the numerator using this identity. then you shud get (2(sinx)^2 + 2sinx)/2sinxcosx...then take out common factor in numerator of 2sinx...then u get 2sinx(sinx+1)/2sinxcosx...now cancel 2sinx and you get (sinx+1)/cosx....its not that long doing it, but typing it out like this makes it seem long
uhm. i didnt get the part after the difference of 2 squares...
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