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February 1, 2015

February 1, 2015

Posted by **Carry** on Monday, February 23, 2009 at 8:45pm.

i just did

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

- Trig check my work please? -
**Reiny**, Monday, February 23, 2009 at 9:44pmno, just test it for some value of x

LS is not equal to your RS

- Trig check my work please? -
**Reiny**, Monday, February 23, 2009 at 10:19pmCarry, 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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