JEE Main 2013 (Offline)Question 1
If $x,y,z$ are in A.P. and $\tan^{-1}x$, $\tan^{-1}y$ and $\tan^{-1}z$ are also in A.P., then :
Inverse Trigonometric Functions
Solution
$2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z\Rightarrow\frac{2y}{1-y^2}=\frac{x+z}{1-xz}=\frac{2y}{1-xz}$, so $y^2=xz$; with $2y=x+z$ this gives $(x-z)^2=0$, hence $x=y=z$.