Sin2x/(sinx) - cos2x/(cos x)?

How do u proof that LHS=RHS

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  • 8 years ago
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    Since cos^2(x)=(1+cos2x)/2 so we obtain cos2x=2cos^(x)-1 also we know sin2x=2sin(x)cos(x) now if we put these things in what u have writen we will have [(2sin(x)cos(x)]/sin(x) - (2cos^(x)-1)/cos(x) = 2cos(x)-2cos(x)+ (1/cos(x))=sec(x)

    Source(s): Me
  • 8 years ago

    Only one side of equation is given

    any how you can simplify this as follows

    sin 2x / sin x - cos 2x / cos x

    Taking LCM we get

    (sin2x cosx - cos 2x sin x) /sin x cos x

    sin(2x -x) / sinx cos x using Sin (A-B) sinAcosB - cosAsinB

    sin x/ sin x cos x = 1/cos x

    sec x

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