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$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:

  1. $x$, $y$, $z$ are in geometric progression
  2. $x$, $y$, $z$ are in harmonic progression.
  3. $x=y=z$
  4. $(x-y)^2 +(y-z)^2+(z-x)^2 =0$

My attempt:

$$\text{A}=\tan^{-1}x \qquad \text{B}=\tan^{-1}y \qquad \text{C}=\tan^{-1}z$$ $$x=\tan A \qquad y=\tan B \qquad z=\tan C $$ $$x+z=2y$$ $$A+C=2B$$ $$\tan(A + B + C)=\frac{\tan A +\tan B +\tan C - \tan A\tan B\tan C }{1-\tan A\tan B -\tan B\tan C -\tan C\tan A}$$ $$\tan(3B)=\frac{x +y +z - xyz }{1-xy -yz -zx}$$

How do I continue from here?

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2 Answers 2

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(Not an answer so much as an elaborate comment ...)

For fun, I made a "trigonograph" to see why we might expect the elements to be equal (item (3)). Here, the arithmetic progressions are $\alpha$, $\gamma$, $\beta$ and $\tan\alpha$, $\tan\delta$, $\tan\beta$. (Also, I take all values to be positive, and assume $\alpha + \beta < 180^\circ$.)

enter image description here

$$\begin{align} 2 \gamma &\;=\; \alpha + \beta \\[4pt] 2 \tan\delta &\;=\; \tan\alpha + \tan\beta \end{align}\qquad\implies\qquad \begin{array}{c} \gamma \;\leq\; \delta \\ \text{with equality when and only when} \\ \alpha = \beta \;\left(\; = \gamma = \delta\;\right) \end{array}$$

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HINT:

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

$$\implies\tan(\tan^{-1}z-\tan^{-1}y)=\tan(\tan^{-1}y-\tan^{-1}x)$$

$$\iff\dfrac{z-y}{1+yz}=\dfrac{y-x}{1+xy}$$

As $x,y,z$ are in A.P.,$y-x=z-y$

Case$\#1:$If If $y-x=z-y=0$

Case$\#2:$ If $y-x=z-y\ne0, 1+yz=1+xy\iff y(z-x)=0$

$\implies y=0$ or $z-x=0\iff z=x$

Can you take it from here?

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