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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 -

Vedclass · Accepted Answer

$1/3$ and $3$

Vedclass · Answer

(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 $t \ge 0$ and $t 
eq 3$ (since $	an 3x$ must be defined).Then $y = \frac{1 - 3t}{3 - t}$.To find the range,we solve for $t$ in terms of $y$:$y(3 - t) = 1 - 3t \implies 3y - yt = 1 - 3t \implies 3t - yt = 1 - 3y \implies t(3 - y) = 1 - 3y$.$t = \frac{1 - 3y}{3 - y} = \frac{3y - 1}{y - 3}$.Since $t = 	an^2 x \ge 0$,we have $\frac{3y - 1}{y - 3} \ge 0$.Using the wavy curve method,the expression is $\ge 0$ when $y \le 1/3$ or $y \ge 3$.Thus,$y$ never lies between $1/3$ and $3$.

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

Similar Questions

If $\frac{\cos 3A + \sin 3A}{\cos A - \sin A} = 1 - K \sin 2A,$ the value of $K$ is

$2\sin^2 \beta + 4\cos(\alpha + \beta)\sin \alpha \sin \beta + \cos 2(\alpha + \beta) = $

If $\alpha = 22^\circ 30'$,then $(1 + \cos \alpha )(1 + \cos 3\alpha )(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ equals

If $\tan^2 \alpha \tan^2 \beta + \tan^2 \beta \tan^2 \gamma + \tan^2 \gamma \tan^2 \alpha + 2 \tan^2 \alpha \tan^2 \beta \tan^2 \gamma = 1$,then the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ is

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$,then the value of $3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module