Question

cot-¹9+cosec-¹√41/4=π/4 prove the statement​

Answer 1

Answer:

${cot}^{ - 1} 9 + {cosec}^{ - 1} \frac{ \sqrt{41} }{4} \\ \\ = {tan}^{ - 1} \frac{1}{9} + {tan}^{ - 1} \frac{ \frac{4}{\sqrt{41} } }{ \sqrt{1 - \frac{16}{41} } } \\ \\ = {tan}^{ - 1} \frac{1}{9} + {tan}^{ - 1} \frac{4}{5} \\ \\ = {tan}^{ - 1} \frac{ \frac{1}{9} + \frac{4}{5} }{1 - \frac{1}{9} \times \frac{4}{5} } \\ \\ = {tan}^{ - 1} \frac{5 + 36}{45 - 4} \\ \\ = {tan}^{ - 1} \frac{41}{41} \\ \\ = {tan}^{ - 1} (1) \\ \\ = \frac{\pi}{4} \\ \\ proved$