The value of frac{tan x}{tan 3x},whenever def | vedclass

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(A) Let $y = \frac{	an x}{	an 3x}$.Using the formula $	an 3x = \frac{3	an x - 	an^3 x}{1 - 3	an^2 x}$,we get:$y = \frac{	an x}{\frac{3	an x -

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$1/3$ and $3$

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(A) Let $y = \frac{	an x}{	an 3x}$.Using the formula $	an 3x = \frac{3	an x - 	an^3 x}{1 - 3	an^2 x}$,we get:$y = \frac{	an x}{\frac{3	an x - 	an^3 x}{1 - 3	an^2 x}} = \frac{1 - 3	an^2 x}{3 - 	an^2 x}$.Let $t = 	an^2 x$,where $t \ge 0$. Then $y = \frac{1 - 3t}{3 - t}$.To find the range,let $y = \frac{3(1 - 3t)}{3(3 - t)} = \frac{3 - 9t}{3(3 - t)} = \frac{(3 - t) - 8t}{3(3 - t)} = \frac{1}{3} - \frac{8t}{3(3 - t)}$.Alternatively,solving for $t$: $y(3 - t) = 1 - 3t \implies 3y - yt = 1 - 3t \implies t(3 - y) = 1 - 3y \implies t = \frac{1 - 3y}{3 - y}$.Since $t = 	an^2 x \ge 0$,we have $\frac{1 - 3y}{3 - y} \ge 0$.This inequality holds when $y \in [1/3, 3)$.Thus,the value of $y$ never lies in the interval $(1/3, 3)$.

The value of $\frac{\tan x}{\tan 3x}$,whenever defined,never lies between

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