Question

If x, y and z are in A.P and tan-1x, tan-1y  and  tan-1zare also in A.P, then

Answer 1

Answer:

$\large\boxed{\sf{{x}^{2}+1=0}}$

Step-by-step explanation:

Given that,

x, y and z are in AP.

Therefore, we will get,

=> x + z = 2y

Also, it's given that,

${\tan}^{-1}x$, ${\tan}^{-1}y$ and ${\tan}^{-1}z$ are in AP.

Therefore, we will get,

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

Now, on both sides, function is same.

Therefore, the domains will also be same.

Therefore, we will get,

$= > \dfrac{x + z}{1 - xz} = \dfrac{2x}{1 - {x}^{2} } \\ \\ = > \dfrac{2y}{1 - xz} = \dfrac{2x}{1 - {x}^{2} } \\ \\ = > y(1 - {x}^{2} ) = x(1 - xz) \\ \\ = > y - y {x}^{2} = x - z {x}^{2} \\ \\ = > y - x = {x}^{2} (y - z) \\ \\ = > y - x = - {x}^{2} (z - y) \\ \\ = >( y - x )= - {x}^{2} (y - x) \\ \\ = > 1 = - {x}^{2} \\ \\ = > {x}^{2} + 1 = 0$

Hence, the required relation is ${x}^{2}+1=0$.