Question

prove this ,,
don't copy from internet please​

Answer 1

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$2 {tan}^{ - 1} \frac{1}{2} + {tan}^{ - 1} \frac{1}{7} \\ = {tan}^{ - 1} \frac{2 \times \frac{1}{2} }{1 - ( { \frac{1}{2} })^{2} } + {tan}^{ - 1} \frac{1}{7} \\ = {tan}^{ - 1} \frac{1}{ \frac{3}{4} } + {tan}^{ - 1} \frac{1}{7} \\ = {tan}^{ - 1} \frac{4}{3} + {tan}^{ - 1} \frac{1}{7} \\ = {tan}^{ - 1} \frac{ \frac{4}{3} + \frac{1}{7} }{1 - \frac{4}{3} \times \frac{1}{7} } \\ = {tan}^{ - 1} \frac{28 + 3}{21 - 4} \\ = {tan}^{ - 1} \frac{31}{17}$