Prove the range for tan3x/tanx is 3 and 1/3

prove that

tan 3x/ tan x never lies in the range3 and 1/3 whenever define!

5 Answers

Anonymous ·

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

1
krish1092 ·

tan3x/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

1
Philip Calvert ·

tan3x= \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}

1
Philip Calvert ·

.......dint see KS's post

1
MATRIX ·

[3][3][3]........Philip..........

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