prove identity sin 2x/1-sinx= secx+1/secx solve sin2y=cos4y for y, where between 0 degree and 360 degree cosx(secx-cscx)= 1-cotx (a) solve 11sin 2 y=13-3sin 2 y for y, where 0 to 360 degree (b) what second-quadrant angle satisfices this equation? (c) what fouth- quadrant angle satisfices this equation? Hi, I assume that L.S. = sin 2x/(1-sinx) and not sin 2x/1-sinx which would be the same as sin 2x - sinx. It seems reasonable to start with the right side by changing secx into cosx and finding a common denominator: R.S. = secx + 1/secx = 1/cosx + cosx = 1/cosx + cos 2x/cosx = (1 + cos 2x)/cosx The denominators of cos x (R.S) and 1-sinx (L.S.) requires a little trick to deal with. cos 2x=1-sin 2x=(1-sinx)(1+sinx) suggests we multiply numerator and denominator of the left side by 1+sinx, and hope things work out (note that this will have the added benefit of getting rid of a difference in the denominator of L.S.): L.S. = (sin 2x)(1+sinx)/(1-sinx)(1+sinx) = (1-cos 2x)(1+sinx)/cos 2x = (1-cos 2x+sinx-sinxcos 2x)/cos 2x I am now starting to think that the identity is false. So I tried 30 degrees. L.S.(x=30) = 1/2 R.S.(x=3-) = 7*sqr(3)/6 Thus, the identity is false. I decided not to spend any time with the remaining questions in case there was a problem with these as well. Paul Go to Math Central