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So the question is $$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$ So my take on the question is to rewrite it as

$$ \tan^{-1}\left(x\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(y\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(z\sqrt{\frac{(x+y+z)}{yzx}}\right) $$

Then say $$\frac{x+y+z}{yzx}= a^2.$$ We get $$ \tan^{-1}\left( \frac{a((x+y+z)-a^2xyz)}{1-a^2(xy+yz+zx)}\right)$$

And since $ (x+y+z) = a^2xyz $ , this is just equal to $\tan^{-1}(0)= 0 $ but the answer given is $\pi.$

Blue
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Mr.HiggsBoson
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  • I suspect that you tried to utilize the formula $\arctan A+\arctan B=\arctan((A+B)/(1-AB))$. If so, can you spot an issue in the following non-identity? $$ \arctan(\sqrt{3})+\arctan(\sqrt{3})=\arctan\left(\frac{2\sqrt{3}}{1-\sqrt{3}^2}\right)=\arctan(-\sqrt{3})=-\arctan(\sqrt{3})$$ and see how the naive use of addition formula can go wrong? – Sangchul Lee Jun 08 '19 at 20:25
  • Anyway, perhaps a neat solution would be employing some cleverly chosen triangle, although I can't easily think of one. Instead, you can check that all the first-order partial derivatives vanish (and of course, it suffices to check for just one partial derivative and then argue by symmetry). From this, we can conclude that the sum is at constant function. Then plugging some nice choices such as $x = y = z = 1$ will determine the value of the constant. – Sangchul Lee Jun 08 '19 at 20:28
  • $\tan\theta = 0$ doesn't imply $\theta = 0$. Instead, it implies $\theta = N\pi$ for some integer $N$. You are summing three $\tan^{-1}\theta_i \in (0,\frac{\pi}{2})$, the sum $\in (0,\frac{3\pi}{2})$ and so $N$ can only be $1$. – achille hui Jun 09 '19 at 05:25

4 Answers4

3

Let $x$, $y$ and $z$ be positive numbers. We consider a triangle $ABC$ with side lengths $a=BC=y+z$, $b=CA=x+z$ and $c=AB=x+y$. The semi-perimeter $s=x+y+z$ inradius $r$. Now, by Heron’s formula we have $$\eqalign{\cot(A/2)&=\frac{s-a}{r}=\frac{s(s-a)}{{\rm Area}(ABC)}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\cr &=\sqrt{\frac{x(x+y+z)}{yz}}}$$ So, $$\eqalign{\tan^{-1}\sqrt{\frac{x(x+y+z)}{yz}}&=\frac{\pi-A}{2}\cr \tan^{-1}\sqrt{\frac{y(x+y+z)}{zx}}&=\frac{\pi-B}{2}\cr \tan^{-1}\sqrt{\frac{z(x+y+z)}{xy}}&=\frac{\pi-C}{2}}$$ Adding we get $\pi$ as a sum.

Omran Kouba
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Use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values,

$-\dfrac\pi2\le \tan^{-1}a\le\dfrac\pi2$

Now $\sqrt b\ge0$ for real $\sqrt b$

$\implies0\le\tan^{-1}\sqrt b\le\dfrac\pi2$

So, here the sum will lie in $\in[0,3\pi/2]$

Now the sum will be $=0$ only if each term under is individually $=0$

i.e. if $x+y+z=0$

Otherwise the sum will be $\ne0$

Also, the general value of the sum is $n\pi$ where $n$ is an integer

So, for $x+y+z\ne0,n=1$

lab bhattacharjee
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Hint:- $$ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c= \pi$$ Only and only if

$$a+b+c=abc$$

Abhishek Kumar
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0

Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$\tan^{-1}\sqrt{\dfrac{x(x+y+z)}{yz}}+\tan^{-1}\sqrt{\dfrac{y(x+y+z)}{zx}}$$ $$=\begin{cases} \tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}\right) &\mbox{if } \sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}<1 \ \ \ \ (1) \\ \pi+\tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}\right) & \mbox{if } \sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}>1 \ \ \ \ (2) \end{cases} $$

Now $R=\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}=\sqrt{\dfrac{x+y+z}{xyz}}\cdot\dfrac{|z|(|x|+|y|)}{|z|-|x+y+z|}$

If $|x+y|,|z|,|x+y+z|\ge0,$ $$R=-\sqrt{\dfrac{x+y+z}{xyz}}\cdot z=-\sqrt{\dfrac{z(x+y+z)}{xy}}$$

Again $(2)$ will hold true if $\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}>1 \iff (x+y+z)^2>z^2$

which is true if $x,y,z>0$

Finally $\tan^{-1}(-u)=-\tan^{-1}u$

lab bhattacharjee
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