Question

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

• A. $x = y = z$
• B. $xy = yz$
• C. $x^2 = yz$
• D. $z^2 = xy$

Answer 1

Answer: (A)

$x = y = z$

Answer 2

Since $x, y, z$ are in $A.P .$ $\therefore 2y = x+z \quad...\left(1\right)$ Since $tan^{-1}x, tan^{-1}y, tan^{-1}z$ and in $A.P.$ $\therefore 2tan^{-1}y = tan^{-1}x+tan^{-1}z$ $tan^{-1} \frac{2y}{1-y^{2}} = tan^{-1}\frac{ x+z}{1-xz}$ $\Rightarrow \frac{2y}{1-y^{2}} =\frac{ x+z}{1-xz}$ $\therefore x, y, z$ are in $H.P.$ Also $y^{2} = xz$ $\Rightarrow x, y, z$ are in $G.P.$ $\therefore x= y=z$