Question

please solve this problem​

Answer 1

LHS:

${tan}^{ - 1} \frac{3}{4} + {tan}^{ - 1} \frac{3}{5} - {tan}^{ - 1} \frac{8}{19} \\ = {tan}^{ - 1} ( \frac{ \frac{3}{4} + \frac{3}{5} }{1 - \frac{3}{4} .\frac{3}{5} } ) - {tan}^{ - 1} \frac{8}{19} \\ = {tan}^{ - 1} ( \frac{15 + 12}{20 - 9} ) - {tan}^{ - 1} \frac{8}{19} \\ = {tan}^{ - 1} \frac{27}{11} - {tan}^{ - 1} \frac{8}{19} \\ = {tan}^{ - 1} ( \frac{ \frac{27}{11} - \frac{8}{19} }{1 + \frac{27}{11}. \frac{8}{19} } ) \\ = {tan}^{ - 1} ( \frac{513 - 88}{209 + 216} ) \\ = {tan}^{ - 1} \frac{425}{425} \\ = {tan}^{ - 1} 1 \\ = \frac{\pi}{4}$

Answer 2

Step-by-step explanation:

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