Simplified form of $$\tan \left( {\frac{\pi }{4} + \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right) + \tan \left( {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right){\text{ is}}$$

Solution :
$$\eqalign{ & {\text{Let}}\frac{1}{2}{\cos ^{ - 1}}\frac{a}{b} = \theta ;{\text{ then}} \cr & {\cos ^{ - 1}}\frac{a}{b} = 2\theta ; \cr & \Rightarrow \cos 2\theta = \frac{a}{b}{\text{ then expression}} \cr & = \tan \left( {\frac{\pi }{4} + \theta } \right) + \tan \left( {\frac{\pi }{4} - \theta } \right) \cr & = \frac{{1 + \tan \theta }}{{1 - \tan \theta }} + \frac{{1 - \tan \theta }}{{1 + \tan \theta }} \cr & = \frac{{{{\left( {1 + \tan \theta } \right)}^2} + {{\left( {1 - \tan \theta } \right)}^2}}}{{\left( {1 - \tan \theta } \right)\left( {1 + \tan \theta } \right)}} \cr & = \frac{{2 + 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }} = \frac{{2\left( {1 + {{\tan }^2}\theta } \right)}}{{1 - {{\tan }^2}\theta }} \cr & = \frac{{2\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}}{{\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}} \cr & = \frac{2}{{\cos 2\theta }} = \frac{2}{{\frac{a}{b}}} = \frac{{2b}}{a} \cr} $$