Question

Question 8: Prove tan¯¹ 1/5 + tan¯¹ 1/7 + tan¯¹ 1/3 + tan¯¹ 1/8 = π/4

Class 12 - Math - Inverse Trigonometric Functions

Answer 1

⇒ tan¯¹ 1/5 + tan¯¹ 1/7 + tan¯¹ 1/3 + tan¯¹ 1/8 = π/4
⇒ tan¯¹ [ (1/5 + 1/7) / (1- 1/5×1/7) ] +  tan¯¹ [ (1/3 + 1/8) / (1 - 1/3×1/8) ]
⇒ tan¯¹ [ (7+5) / (35 -1) ] + tan¯¹ [ (8+3) / (24 -1) ]
⇒ tan¯¹ (12/34) + tan¯¹ (11 /23)
⇒ tan¯¹ (6/17) + tan¯¹ (11 /23)
⇒ tan¯¹ [ (6/17 + 11/23) / (1- 6/17×11/23) ]
⇒ tan¯¹  [ (138 + 187) / (391 - 66) ]
⇒ tan¯¹ (325 / 325)
⇒ tan¯¹ (1)
⇒ π/4

Hope it helps!!!

Answer 2

${tan}^{ - 1} \frac{1}{5} + {tan}^{ - 1} \frac{1}{7} + {tan}^{ - 1} \frac{1}{3} + {tan}^{ - 1} \frac{1}{8} \\ \\ = {tan}^{ - 1} ( \frac{ \frac{1}{5} + \frac{1}{7} }{1 - \frac{1}{5} . \frac{1}{7} } ) + {tan}^{ - 1} ( \frac{ \frac{1}{3} + \frac{1}{ 8 } }{1 - \frac{1}{3}. \frac{1}{8} } ) \\ \\ = {tan}^{ - 1} ( \frac{7 + 5}{35 - 1} ) + {tan}^{ - 1} ( \frac{8 + 3}{24 - 1} ) \\ \\ = {tan}^{ - 1} ( \frac{12}{34} ) + {tan}^{ - 1} ( \frac{11}{23} ) \\ \\ = {tan}^{ - 1} \frac{6}{17} + {tan}^{ - 1} \frac{11}{23} \\ \\ = {tan}^{ - 1} ( \frac{ \frac{6}{17} + \frac{11}{23} }{1 - \frac{6}{17} . \frac{11}{23} } ) \\ \\ = {tan}^{ - 1} ( \frac{138 + 187}{391 - 66} ) \\ \\ = {tan}^{ - 1} ( \frac{325}{325} ) \\ \\ = {tan}^{ - 1} 1 \\ \\ = \frac{\pi}{4}$