Trig check my work please?

tan(X+30)tan(30-x)=2cos2x-1 / (2cos2x+1)

i just did

(tanx)(-tanx)=-sin^2x/cos^2x would that work?

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  1. no, just test it for some value of x
    LS is not equal to your RS

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  2. Carry, where are you getting these,
    even though I enjoy doing them, they are getting to me, lol

    recall tan(A+B) = (tanA + tanB)/(1 - tanAtanB)

    so LS
    = (tanx + tan30)/1-tanxtan30)*(tan30-tanx)/(1+tanxtan30)
    = (tan^2 30 - tan^2 x)/(1 - (tan^x)(tan^2 30))
    now tan^2 30º = 1/3
    so the above
    = (1/3 - tan^2 x)/(1 - (1/3)tan^2 x)
    multiply top and bottom by 3 to get
    (1 - 3tan^2 x)/(3 - tan^2 x)

    RS = (2cos(2x) - 1)/(2cos(2x) + 1)
    = (2(cos^2x - sin^2x) - 1)/(2(cos^2x-sin^2x) + 1)
    remember 1 = sin^2x + cos^2x , so we get
    (2cos^2x - 2sin^x - sin^2x - cos^2x)/(2cos^2x - 2sin^2x + sin^2x + cos^2x)
    = (cos^2x - 3sin^2x)/(3cos^2x - sin^2x)
    divide top and bottom by cos^x to get
    (1 - 3tan^2x)/(3 - tan^2x)

    ok, no more!!!!

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