no, just test it for some value of x
LS is not equal to your RS
tan(X+30)tan(30-x)=2cos2x-1 / (2cos2x+1)
i just did
(tanx)(-tanx)=-sin^2x/cos^2x would that work?
2 answers
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!!!!
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!!!!
2 answers