If x, y, z are in A.P. and tan ^{-1} x, tan ^ | 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.$$	herefore 2y = x + z$ ... $(i)$Also,$	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z$ are in $A.P.$$	herefore 2 	an

Vedclass · Accepted Answer

$x=y=z$

Vedclass · Answer

(A) Given that $x, y, z$ are in $A.P.$$	herefore 2y = x + z$ ... $(i)$Also,$	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z$ are in $A.P.$$	herefore 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)$:$	an ^{-1} \left( \frac{2y}{1-y^2} ight) = 	an ^{-1} \left( \frac{x+z}{1-xz} ight)$$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$Substituting $x+z = 2y$ from $(i)$:$\frac{2y}{1-y^2} = \frac{2y}{1-xz}$This implies either $2y = 0$ (which leads to $x=y=z=0$) or $1-y^2 = 1-xz$,which gives $y^2 = xz$.Since $x, y, z$ are in both $A.P.$ and $G.P.$,we must have $x=y=z$.

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 $y = \sec^{-1}\left( \frac{x + 1}{x - 1} \right) + \sin^{-1}\left( \frac{x - 1}{x + 1} \right)$,then $\frac{dy}{dx} = $

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals

If $y = \tan^{-1} \frac{x}{1+2x^2} + \tan^{-1} \frac{x}{1+6x^2} + \tan^{-1} \frac{x}{1+12x^2}$,then $\left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = $

Prove $\cos ^{-1} \frac{4}{5} + \cos ^{-1} \frac{12}{13} = \cos ^{-1} \frac{33}{65}$

If $\pi \leq x \leq 2 \pi$,then $\cos^{-1}(\cos x)$ is equal to :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module