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

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

Question

(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$. Since $	an^{-1}x, 	an^{-1}y, 	an^{-1}z$ are in $A.P.$,we have $2	an^{-1}y = 	an^{-1}

Vedclass · Accepted Answer

$x = y = z$

Vedclass · Answer

(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$. Since $	an^{-1}x, 	an^{-1}y, 	an^{-1}z$ are in $A.P.$,we have $2	an^{-1}y = 	an^{-1}x + 	an^{-1}z$. Using 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 $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$. Substituting $x+z = 2y$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$. This equation holds if $2y = 0$ or $1-y^2 = 1-xz$. If $2y = 0$,then $y=0$,which implies $x+z=0$ or $x=-z$. If $1-y^2 = 1-xz$,then $y^2 = xz$. Since $x, y, z$ are in $A.P.$ ($y^2 = xz$ implies $G.P.$),the only way for numbers to be in both $A.P.$ and $G.P.$ is if $x = y = z$.

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

Similar Questions

If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$,then for $x < 0$ the value of $500 x^4+270 x^2+997=$

All $x$ satisfying the inequality $(\cot^{-1} x)^2 - 7(\cot^{-1} x) + 10 > 0$ lie in the interval

The number of solutions of $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = 0$ is

Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \csc x)$

$\cot \left\{\frac{2019 \pi}{2}-\left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)\right\}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module