Question

Show that the value of
tanx/tan 3x, wherever defined, never lies between 1/3 and 3.
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{tanx}{tan3x}$

Let assume that

$\rm :\longmapsto\:y = \dfrac{tanx}{tan3x}$

We know that

$\rm :\longmapsto\:tan3x = \dfrac{3tanx - {tan}^{3}x}{1 - 3 {tan}^{2}x}$

So, on substituting this value, we get

$\rm :\longmapsto\:y = \dfrac{tanx}{\dfrac{3tanx - {tan}^{3}x}{1 - {3tan}^{2} x} }$

$\rm :\longmapsto\:y = \dfrac{tanx(1 - {3tan}^{2} x)}{tanx(3 - {tan}^{2}x)}$

$\rm :\longmapsto\:y = \dfrac{1 - {3tan}^{2} x}{3 - {tan}^{2}x}$

$\rm :\longmapsto\:3y - y {tan}^{2}x = 1 - {3tan}^{2}x$

$\rm :\longmapsto\:3 {tan}^{2}x - y {tan}^{2}x = 1 - 3y$

$\rm :\longmapsto\:y{tan}^{2}x - 3 {tan}^{2}x = 3y - 1$

$\rm :\longmapsto\:{tan}^{2}x(y - 3 ) = 3y - 1$

$\rm :\longmapsto\: {tan}^{2}x = \dfrac{3y - 1}{y - 3}$

$\bf\implies \:tanx = \sqrt{\dfrac{3y - 1}{y - 3} }$

Now, tanx is defined iff

$\rm :\longmapsto\:\dfrac{3y - 1}{y - 3} \geqslant 0 \: \: and \: y \: \ne \: 3$

$\rm :\longmapsto\:\dfrac{(3y - 1)(y - 3)}{(y - 3)^{2} } \geqslant 0 \: \: and \: y \: \ne \: 3$

$\rm :\longmapsto\:(3y - 1)(y - 3)\geqslant 0 \: \: and \: y \: \ne \: 3$

$\bf\implies \:y \leqslant \dfrac{1}{3} \: \: or \: \: y > 3$

$\bf\implies \:\dfrac{tanx}{tan3x} \leqslant \dfrac{1}{3} \: \: \: \: or \: \: \: \dfrac{tanx}{tan3x} > 3$

Hence, Proved

Additional Information :-

If a and b are two positive real numbers such that a < b, then

$\boxed{ \sf{ \:(x - a)(x - b) < 0 \: \rm \implies\:a < x < b}}$

$\boxed{ \sf{ \:(x - a)(x - b) \leqslant 0 \: \rm \implies\:a \leqslant x \leqslant b}}$

$\boxed{ \sf{ \:(x - a)(x - b) > 0 \: \rm \implies\: x < a \: \: or \: x > b}}$

$\boxed{ \sf{ \:(x - a)(x - b) \geqslant 0 \: \rm \implies\: x \leqslant a \: \: or \: x \geqslant b}}$

$\boxed{ \sf{ \: |x| < y\rm \: \implies\: - y < x < y}}$

$\boxed{ \sf{ \: |x| \leqslant y\rm \: \implies\: - y \leqslant x \leqslant y}}$

$\boxed{ \sf{ \: |x| > y\rm \: \implies\: x < - y \: \: or \: \: x > y}}$

$\boxed{ \sf{ \: |x| \geqslant y\rm \: \implies\: x \leqslant - y \: \: or \: \: x \geqslant y}}$