Example 11 - Show sin-1 12/13 + cos-1 4/5 + tan-1 63/16 = pi - Examples

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
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Example 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - NCERT Book (Method 1) Show that sin-1 12/13 + cos-1 4/5 + tan-1 63/16 = Ο€ Let a = sin-1 12/13 and b = cos-1 4/5 We convert sin-1 & cos-1 to tan–1 & then use tan (a + b) formula Let a = sin-1 12/13 sin a = 12/13 We know that cos2 a = 1 – sin2 a cos a = √(1 βˆ’sin2β‘π‘Ž ) =√(1βˆ’(12/13)^2 ) " =" √(1βˆ’144/169) " =" √((169 βˆ’144)/169) " =" √(25/169) " =" √(52/132) "=" 5/13 Let tan a = sinβ‘π‘Ž/cosβ‘π‘Ž = (12/13)/(5/13) = 12/13 Γ— 13/5 = 12/5 tan a = 12/5 Let b = cos-1 4/5 cos b = 4/5 We know that sin2 b = 1 – cos2 b sin b = √("1 – cos2 b " ) = √("1 βˆ’ " (4/5)^2 ) = √(1 βˆ’16/25) = √((25 βˆ’ 16)/25) = √(9/25) = 3/5 Let tan b = sin⁑𝑏/cos⁑𝑏 = (3/5)/(4/5) = 3/5 Γ— 5/4 = 3/4 tan b = 3/4 We know that tan (a + b) = 𝒕𝒂𝒏⁑〖𝒂 +γ€– 𝒕𝒂𝒏〗⁑〖𝒃 γ€— γ€—/(𝟏 βˆ’ 𝒕𝒂𝒏⁑〖𝒂 𝒕𝒂𝒏⁑𝒃 γ€— ) Putting tan a = 12 and tan b = 3/4 = (12/5 + 3/4)/(1 βˆ’ 12/5 Γ— 3/4) = ((48 +15 )/20)/((20 βˆ’ 36)/20) = (63/20)/((βˆ’16)/20) = 63/20 Γ— 20/(βˆ’16) = βˆ’63/( 16) Hence, tan (a + b) = βˆ’ 63/16 a + b = tan-1 (βˆ’( 63)/16) Putting value of a & b sin-1 12/13 + cos-1 4/5 = tan-1 (βˆ’( 63)/16) Taking L.H.S sin-1 12/13 + cos-1 4/5 + tan-1 63/16 Putting values = tan-1 (βˆ’63/16) + tan-1 (63/16) Using tan-1x + tan-1y = tan-1((𝒙 + π’š)/(𝟏 βˆ’ π’™π’š)) Putting x = (βˆ’ 63)/16 and y by = 63/16 = tan-1(((βˆ’ 63)/16 + 63/16)/(1 βˆ’ (βˆ’ 63)/16 Γ— 63/16)) = tan-1(0/(1 βˆ’ (( 63)/16)^2 )) As tan 180 = 0 tan Ο€ = 0 Ο€ = tan-1 0 i.e. tan-1 0 = Ο€ = tan-1 0 = Ο€ = R.H.S Hence L.H.S = R.H.S Hence proved Example 11 (Method 2) Show that sin-1 12/13 + cos-1 4/5 + tan-1 63/16 = Ο€ Let a = sin-1 12/13 and b = cos-1 4/5 We convert sin-1 & cos-1 to tan–1 & then use tan-1 a + tan-1b formula Let a = sin-1 12/13 sin a = 12/13 We know that cos2 a = 1 – sin2 a cos a = √(1 βˆ’sin2β‘π‘Ž ) =√(1βˆ’(12/13)^2 ) " =" √(1βˆ’144/169) " =" √((169 βˆ’144)/169) " =" √(25/169) " =" √(52/132) "=" 5/13 Let tan a = sinβ‘π‘Ž/cosβ‘π‘Ž = (12/13)/(5/13) = 12/13 Γ— 13/5 = 12/5 tan a = 12/5 a = tan-1 𝟏𝟐/πŸ“ Let b = cos-1 4/5 cos b = 4/5 We know that sin2 b = 1 – cos2 b sin b = √("1 – cos2 b " ) = √("1 βˆ’ " (4/5)^2 ) = √(1 βˆ’16/25) = √((25 βˆ’ 16)/25) = √(9/25) = 3/5 Let tan b = sin⁑𝑏/cos⁑𝑏 = (3/5)/(4/5) = 3/5 Γ— 5/4 = = 3/4 tan b = 3/4 b = tan-1 πŸ‘/πŸ’ Taking L.H.S sin-1 12/13 + cos-1 4/5 + tan-1 63/16 = a + b + tan-1 63/16 Putting values = tan-1 12/5 + tan-1 3/4 + tan-1 63/16 Using tan-1x + tan-1y = tan-1((𝒙 + π’š)/(𝟏 βˆ’ π’™π’š)) Putting x = 12/5 and y = 3/4 = tan-1 ((12/5 + 3/4)/(1 βˆ’ 12/5 Γ— 3/4))+ tan-1 63/16 = tan-1 (((12 Γ— 4 + 3 Γ— 5 )/20)/(1 βˆ’ 36/20))+ tan-1 63/16 = tan-1 (((48 +15 )/20)/((20 βˆ’ 36)/20))+ tan-1 63/16 = tan-1 ((63/20)/((βˆ’16)/20))+ tan-1 63/16 = tan-1 (63/20Γ—20/(βˆ’16))+ tan-1 63/16 = tan-1 (βˆ’ 63/( 16)) + tan-1 63/16 Using tan-1x + tan-1y = tan-1((𝒙 + π’š)/(𝟏 βˆ’ π’™π’š)) Putting x = (βˆ’ 63)/16 and y by = 63/16 = tan-1(((βˆ’ 63)/16 + 63/16)/(1 βˆ’ (βˆ’ 63)/16 Γ— 63/16)) = tan-1(((βˆ’ 63)/16 + 63/16)/(1 βˆ’ (βˆ’ 63)/16 Γ— 63/16)) = tan-1(0/(1 βˆ’ (( 63)/16)^2 )) As tan 180Β° = 0 tan Ο€ = 0 Ο€ = tan-1 0 i.e. tan-1 0 = Ο€ = tan-1 0 = Ο€ = R.H.S Hence L.H.S = R.H.S , Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.