Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$

Question

The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to

$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{12} $

$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$ $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$ $$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$

Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$ How to solve this ?

From where are you getting these problems ? which book ? – Jul 31 '16 at 06:30 — , Jul 31 '16 at 06:30
@ritwiksinha I got this in test. – Aakash Kumar Jul 31 '16 at 06:33 — Aakash Kumar, Jul 31 '16 at 06:33

lab bhattacharjee · Accepted Answer · 2016-07-30T11:09:35.700

7

$$\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\dfrac{\sqrt3-\sqrt2}{1+\sqrt3\cdot\sqrt2}$$

$$\implies\arctan\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\arctan\sqrt3-\arctan\sqrt2$$

$$\arctan\sqrt3=\dfrac\pi3$$ and

$$\arctan\sqrt2=\text{arccot}\dfrac1{\sqrt2}=\dfrac\pi2-\arctan\dfrac1{\sqrt2}$$

edited Jul 30 '16 at 11:09

answered Jul 30 '16 at 11:06

lab bhattacharjee

274,582

See also : http://math.stackexchange.com/questions/1430520/proving-that-arctanx-arctan1-x-pm-pi-2-could-this-line-of-reasoning-p and http://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function – lab bhattacharjee Jul 30 '16 at 11:07
@AakashKumar, Sorry for the typo. Please let me know if you have any further query – lab bhattacharjee Jul 30 '16 at 11:10
$$(\sqrt3 -\sqrt2)^2=(\sqrt2 -\sqrt3)^2$$ we should also check for $$(\sqrt2 -\sqrt3)$$ – Aakash Kumar Jul 30 '16 at 11:23
@AakashKumar, Observe that for real $a\ge0$ $$\sqrt{a}\ge0$$ – lab bhattacharjee Jul 30 '16 at 11:26
Still I didn't get you , to which you are referring by $a$ – Aakash Kumar Jul 30 '16 at 11:30
@AakashKumar, I mean $$\sqrt{(\sqrt3-\sqrt2)^2}=|\sqrt3-\sqrt2|=\sqrt3-\sqrt2$$ as $\sqrt3-\sqrt2>0$ – lab bhattacharjee Jul 30 '16 at 11:32

score 3 · Answer 2 · answered Jul 30 '16 at 11:13

By using identities $$\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}$$ and $$\tan^{-1}{\alpha}-\tan^{-1}{\beta}=\tan^{-1}(\frac{\alpha-\beta}{1+\alpha\beta}),$$ after a little calculation, we get $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)=\tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}$$

haqnatural · Answer 3 · 2016-07-30T11:10:29.123

Using the fact that $\tan { \left( \alpha -\beta \right) =\frac { \tan { \alpha -\tan { \beta } } }{ 1+\tan { \alpha \tan { \beta } } } } ,\tan { \left( \tan ^{ -1 }{ \alpha } \right) =\alpha } $$$\tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) =t\ \tan { \left( \tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) \right) =\tan { t } } \ \frac { \tan { \left( \tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) \right) -\tan { \left( \tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) \right) } } }{ 1+\tan { \left( \tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) \right) \tan { \left( \tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) \right) } } } =\tan { t } \ \tan { t } =\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } }{ 1+\frac { 1 }{ \sqrt { 2 } } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) } =\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \sqrt { 3 } -\sqrt { 2 } }{ 1+\sqrt { 6 } } }{ 1+\frac { 1 }{ \sqrt { 2 } } \left( \frac { \sqrt { 3 } -\sqrt { 2 } }{ 1+\sqrt { 6 } } \right) } =\frac { 3 }{ 3\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } \ $$

so

$$\tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) =\frac { \pi }{ 6 } $$

I didn't down vote . Your method was long but it is unique and good . — Aakash Kumar, Jul 30 '16 at 11:37

score 1 · Answer 4 · answered Jul 30 '16 at 10:58

1

Hint the second bracket is $\tan(\frac{\pi}{3}-\tan^{-1}\sqrt{2})$ now can you proceed further

answered Jul 30 '16 at 10:58

Archis Welankar

15,910

It can also be $\tan(\tan^{-1}\sqrt{2}-\frac{\pi}{3})$ .I am stuck at part after this. – Aakash Kumar Jul 30 '16 at 11:02
How can it be?. – Archis Welankar Jul 30 '16 at 11:07
$$(\sqrt3 -\sqrt2)^2=(\sqrt2 -\sqrt3)^2$$ – Aakash Kumar Jul 30 '16 at 11:21

score 1 · Answer 5 · 2016-07-30T11:24:53.890

1

We have for all $x >0$, $\tan^{-1} x + \tan^{-1} \frac1{x} = \pi/2$ (hint: take the derivative of LHS). Hence, the obtained expression is just:

$$\frac{\pi}2 - \tan^{-1} \sqrt 3 = \frac{\pi}2 - \frac{\pi}3 = \frac{\pi}6$$

edited Jul 30 '16 at 11:24

answered Jul 30 '16 at 11:05

You are meaning $x>0$, I suppose. For $x<0$, $\arctan x+\arctan(1/x)=-\pi/2$. – egreg Jul 30 '16 at 11:19
@egreg yes, thank you. A priori, the LHS has separate values on $(-\infty, 0)$ and $(0,\infty)$ because of discontinuity. I should've paid attention to that. – Jul 30 '16 at 11:24

score 1 · Answer 6 · answered Jul 30 '16 at 11:33

One should be careful when using identities related to the arctangent, because $\arctan\tan t=t$ only holds for $-\pi/2<t<\pi/2$. On the other hand, $\tan\arctan s=s$ holds without any condition.

Set $$ \alpha=\arctan\frac{1}{\sqrt{2}}, \qquad \beta=\arctan\frac{\sqrt{5-2\sqrt{6}}}{1+ \sqrt{6}}= \arctan\frac{\sqrt{3}-\sqrt{2}}{1+ \sqrt{6}} $$ Note that $$ 1\cdot(1+\sqrt{6})>\sqrt{2}\,(\sqrt{3}-\sqrt{2}) $$ so $\alpha>\beta$ and $0<\alpha-\beta<\pi/2$. Then $$ \tan(\alpha-\beta)= \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}= \frac{\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{3}-\sqrt{2}}{1+ \sqrt{6}}} {1+\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}-\sqrt{2}}{1+ \sqrt{6}}}= \frac{1}{\sqrt{3}} $$ Since $0<\alpha-\beta<\pi/2$, we can conclude $$ \alpha-\beta=\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{6} $$

Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$

6 Answers6