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

Question

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

Aicharya · Accepted Answer

Given, $x, y, \mathrm{z}$ are in A.P. $	herefore \quad 2 y=x+z... (i)$ Also, $\begin{aligned} & 	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z 	ext { are in A.P. } \ 	herefore \quad & 2 	an ^{-1} y=	an ^{-1} x+	an ^{-1} z \ & \Rightarrow 	an ^{-1}\left(\frac{2 y}{1-y^2}ight)=	an ^{-1}\left(\frac{x+z}{1-x z}ight) \ & \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-x z} \ & \Rightarrow \frac{2 y}{1-y^2}=\frac{2 y}{1-x z}... [From (i)] \ & \Rightarrow 1-y^2=1-x z \ & \Rightarrow y^2=x z \end{aligned}$ $	herefore \quad x, y, \mathrm{z} 	ext { are in G.P. }... (ii)$ From (i) and (ii), we get $x=y=\mathrm{z}$

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

Solution