1. prove identity sin 2x/1-sinx= secx+1/secx

  2. solve sin2y=cos4y for y, where between 0 degree and 360 degree

  3. cosx(secx-cscx)= 1-cotx

  4. (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?


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.


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