Question

Prove that:

(tan 3x)/(tan x) lies between 1/3 and 3?

I need ans with steps plz

Answer 1

SOLUTION:-

$\frac{tan3x}{tanx} = \frac{3tan - {tan}^{3} x}{tanx(1 - 3 {tan}^{2} x)} \\ \\ = > \frac{3 - {tan}^{2}x }{1 - 3 {tan}^{2}x }$

Now,

$let \: \frac{3 - {tan}^{2}x }{ 1 - 3 {tan}^{2} x} = a \\ \\ = > 3 - {tan}^{2} x = a - 3a {tan}^{2} x \\ \\ = > {tan}^{2} x(3a - 1) = a - 3 \\ \\ = > {tan}^{2} x = \frac{a - 3}{3a - 1}$

It means,

$\frac{a - 3}{3a - 1} > 0. \\ so ,\: a - 3 > 0 \: \: and \: 3a - 1 < 0 \\ \\ = > a > 3 \: \: and \: \: a < \frac{1}{3} \\ \\ = > so ,\: \frac{tan3x}{tanx} \: never \:lies \: between \: \frac{1}{3} \: \: and \: \: 3.$

Hope it helps ☺️

Answer 2

Answer:

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