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

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

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$x = y = z$

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(A) Given $x, y, z$ are in $A.P.$,we have $2y = x + z$  ....$(1)$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 $2 	an^{-1} y = 	an^{-1} \left( \frac{x+z}{1-xz} ight)$.Taking $	an$ on both sides,$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.Substituting $x+z = 2y$ from $(1)$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.This implies either $2y = 0$ (which leads to $x=y=z=0$) or $\frac{1}{1-y^2} = \frac{1}{1-xz}$,which means $y^2 = xz$.If $x, y, z$ are in both $A.P.$ and $G.P.$,then $x = y = z$.

List $I$	List $II$
$P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value	$1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$
$Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is	$2$. $\sqrt{2}$
$R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is	$3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is	$4$. $1$

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

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