If-the-non-zero-numbers-x-y-z-are-in-AP-and-tan-1-x-tan-1-y-tan-1-z-are-also-in-AP-then-

Question Number 53924 by F_Nongue last updated on 27/Jan/19

If the non - zero numbers x, y, z are in AP,

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

then

Answered by $@ty@m last updated on 27/Jan/19

\tan^{- 1} y - \tan^{- 1} x = \tan^{- 1} z - \tan^{- 1} y

\Rightarrow \tan^{- 1} \frac{y - x}{1 + y x} = \tan^{- 1} \frac{z - y}{1 + z y}

\Rightarrow \frac{y - x}{1 + y x} = \frac{z - y}{1 + z y} \Rightarrow

\Rightarrow 1 + y x = 1 + z y (∵ x, y, z are in AP)

\Rightarrow x = z

\Rightarrow y = z

\Rightarrow x = y = z

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19

g i v e n 2 y = x + z

2 {t a n}^{- 1} y = {t a n}^{- 1} x + {t a n}^{- 1} z

{t a n}^{- 1} (\frac{2 y}{1 - y^{2}}) = {t a n}^{- 1} (\frac{x + z}{1 - x z})

\frac{2 y}{1 - y^{2}} = \frac{x + z}{1 - x z}

2 y - 2 x y z = x + z - {x y}^{2} - {z y}^{2}

{x y}^{2} + {z y}^{2} - 2 x y z = 0

x y + z y - 2 x z = 0

\frac{1}{z} + \frac{1}{x} = \frac{2}{y}

x, y a n d z i n H . P a l s o \dots

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p l s c l e a r l y m e n t i o n i n q u e s t i o n w h a t t o p r o v e \dots