Question

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

(1) $x=y=z$

(2) $2 x=3 y=6 z$

(3) $6 x=3 y=2 z$

Answer 1

SOLUTION : $2 y=x+z  \Rightarrow \quad 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}(z) \\$

$\tan ^{-1}\left(\frac{2 y}{1-y^2}\right)=\tan ^{-1}\left(\frac{x+z}{1-x z}\right)  \Rightarrow \quad \frac{x+z}{1-y^2}=\frac{x+z}{1-x z} \\$

$\Rightarrow \quad y^2=x z \text { or } x+z=0 \Rightarrow  x=y=z$

Above Correct Answer is Option (1)