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 $x \cos \theta - \sin \theta = 1$,then $x^2 - (1 + x^2) \sin \theta$ equals

If $x + y = 3 - \cos 4\theta$ and $x - y = 4 \sin 2\theta$,then:

The value of $\frac{1}{4} \tan \frac{\pi}{8} + \frac{1}{8} \tan \frac{\pi}{16} + \frac{1}{16} \tan \frac{\pi}{32} + \dots \infty$ terms is equal to-

If $\tan \alpha = \frac{m}{m+1}$ and $\tan \beta = \frac{1}{2m+1}$,then $\alpha + \beta =$

If $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3},$ the value of $A + B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module