trigonometry - 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

1. $\frac{\pi}{6}$


2. $\frac{\pi}{4}$


3. $\frac{\pi}{3}$


4. $\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 ?

Answer

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

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

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

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