Discussion Forum

 if x, y and z are  in AP and tan^-1x, tan^-1y and tan^-1 z are also in AP, then

Answer:

Explanation:

 Since x,y and z are in AP

 $\therefore$              2y=x+z

 also  tan^-1x, tan^-1y and tan^-1 z  are in AP

 $\therefore$  2 tan^-1y=tan^-1x= tan^-1 (z)

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

$\Rightarrow$     $\left(\frac{x+z}{1-y^{2}}\right)= \left(\frac{x+z}{1-xz}\right)\Rightarrow y^{2}=xz$  Since x,y and z are in AP as well as in GP

    $\therefore$   x=y=z