1tan3x/tanx=(3-tan^2(x))(1-3tan^2(x))=y (say)
==> (1-3y)tan^2(x)+(y-3)=0
Discriminant≥0
==> y doesn't lie between 1/3 and 3
prove that
tan 3x/ tan x never lies in the range3 and 1/3 whenever define!
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5 Answers
1
1tan3x= \frac{3sinx - 4sin^{3}x}{4cos^{3}x - 3cosx}
=\frac{3sinxcos^{2}x - sin^{3}x}{cos^{3}x-3cosxsin^{2}x} =\frac{3tanx- tan^{3}x}{1-3tan^{2}x}
sorry dint remember this "well known formula"
\Rightarrow tan3x/tanx = \frac{3-tan^{2}x}{1-3tan^{2}x}
Let y= tan3x/tanx
==> y = {3-(tanx)^2}/{1-3(tanx)^2}
{Since, ||tan3x = [tanx {3 - (tanx)^3}]/ 1-3(tanx)^2||}
Solving, (tanx)^2 = (y-3)/(3y-1)
Now, (tanx)^2 >=0 for all x lying in R
==> (y-3)/(3y-1) >= 0
Therefore, y won't lie between 1/3 and 3