Question

Prove that $tan^{-1}(\frac{3}{4}) + tan^{-1}(\frac{3}{5}) - tan^{-1}(\frac{8}{19})= \frac{\pi}{4}$.

Answer 1

As we know that
${tan}^{ - 1} x + {tan}^{ - 1} y = {tan}^{ - 1} ( \frac{x + y}{1 - xy} ) \\ \\ {tan}^{ - 1} ( \frac{3}{4}) + {tan}^{ - 1} ( \frac{3}{5} )= {tan}^{ - 1} ( \frac{ \frac{3}{4} + \frac{3}{5} }{1 - \frac{9}{20} } ) \\ \\ \\ = {tan}^{ - 1} ( \frac{27}{11} )$
Now

${tan}^{ - 1} ( \frac{27}{11}) - {tan}^{ - 1} ( \frac{8}{19} )= {tan}^{ - 1} ( \frac{ \frac{27}{11} - \frac{8}{19} }{1 + \frac{27}{11} \frac{8}{19} } ) \\ \\ \\ = {tan}^{ - 1} ( \frac{27 \times 19 - 88}{11 \times 19 + 27 \times 8}) \\ \\ ={tan}^{-1}(\frac{425}{425}) \\\\\\={tan}^{ - 1} (1) \\ \\ = {tan}^{ - 1} (tan \: \frac{\pi}{4} ) \\ \\ = \frac{\pi}{4}$
hence proved