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 $2 \sin ^{-1} x-3 \cos ^{-1} x=4, x \in[-1,1]$,then $2 \sin ^{-1} x+3 \cos ^{-1} x$ is equal to

If $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4},$ then find the value of $x$.

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$,where $-1 < x < 1$.

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module