x, y, z are in G.P. and tan^{-1} x, tan^{-1} | vedclass

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

Question

(D) Given that $x, y, z$ are in $G$.$P$.,we have $y^2 = xz$ $(i)$.Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A$.$P$.,so $2 	an^{-1} y = 	a

Vedclass · Accepted Answer

$x = y = z$

Vedclass · Answer

(D) Given that $x, y, z$ are in $G$.$P$.,we have $y^2 = xz$ $(i)$.Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A$.$P$.,so $2 	an^{-1} y = 	an^{-1} x + 	an^{-1} z$.Applying the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$,we get $	an^{-1} \left( \frac{2y}{1-y^2} ight) = 	an^{-1} \left( \frac{x+z}{1-xz} ight)$.This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.Since $y^2 = xz$,the denominators are equal,so $2y = x+z$ (ii).From $(i)$ and (ii),$x, y, z$ are in both $A$.$P$. and $G$.$P$.,which implies $x = y = z$.

$x, y, z$ are in $G$.$P$. and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A$.$P$.,then

Similar Questions

If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the incorrect option is

If $x+iy = \frac{1+7i}{(2-i)^2}$,then $\operatorname{cosec}\left(\tan^{-1} \frac{y}{x} - \frac{\pi}{4}\right) = $

The number of positive integral solutions of $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ are

$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

The number of real solutions of $\tan^{-1}\sqrt{x(x + 1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module