Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ 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 ?
 A: $$\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\dfrac{\sqrt3-\sqrt2}{1+\sqrt3\cdot\sqrt2}$$
$$\implies\arctan\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\arctan\sqrt3-\arctan\sqrt2$$
$$\arctan\sqrt3=\dfrac\pi3$$ and
$$\arctan\sqrt2=\text{arccot}\dfrac1{\sqrt2}=\dfrac\pi2-\arctan\dfrac1{\sqrt2}$$
A: 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 $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)=\tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}$$ 
A: 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 

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

A: Hint the second bracket is $\tan(\frac{\pi}{3}-\tan^{-1}\sqrt{2})$ now can you proceed further
A: 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:
$$\frac{\pi}2 - \tan^{-1} \sqrt 3 = \frac{\pi}2 - \frac{\pi}3 = \frac{\pi}6$$
A: 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
$$
\alpha-\beta=\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{6}
$$
