<s>[INST] 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 ?
 [/INST] 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 

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

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