50 pts !!! Help please

[tex]INVERSE \: \: TRIGONOMETRY \\ \\ \\ Given \: \: expression \: - \\ \\ {tan}^{ - 1} \frac{1}{3} + {tan}^{ - 1} \frac{1}{5} + {tan}^{ - 1} \frac{1}{7} + {tan}^{ - 1} \frac{1}{8} \\ \\ \\ Using \: \: the \: \: identity \: \: - \\ {tan}^{ - 1} x + {tan}^{ - 1} y = {tan}^{ - 1} \frac{x + y}{1 - xy} \\ \\ \\ {tan}^{ - 1} \frac{1}{3} + {tan}^{ - 1} \frac{1}{5} + {tan}^{ - 1} \frac{1}{7} + {tan}^{ - 1} \frac{1}{8} \\ \\ = {tan}^{ - 1} (\frac{ \frac{8}{15} }{1 - \frac{1}{15} } ) + {tan}^{ - 1} ( \frac{ \frac{15}{56} }{1 - \frac{1}{56} } ) \\ \\ = {tan}^{ - 1} \frac{8}{14} + {tan}^{ - 1} \frac{15}{55} \\ \\ = {tan}^{ - 1} ( \frac{ \frac{4}{7} + \frac{3}{11} }{1 - \frac{4}{7} \times \frac{3}{11} } ) \\ \\ = {tan}^{ - 1} \frac{65}{65} = {tan}^{ - 1} 1 \\ \\ \\ = \: \: \: \frac{\pi}{4} \: \: \: \: \: \: \: \: \: \: \: Ans.[/tex]