Last updated at Sept. 24, 2018 by Teachoo

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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 tanβ1 & then use tan (a + b) formula Let a = sin-1 12/13 sin a = 12/13 We know that cos2 a = 1 β sin2 a cos a = β(1 βsin2β‘π ) =β(1β(12/13)^2 ) " =" β(1β144/169) " =" β((169 β144)/169) " =" β(25/169) " =" β(52/132) "=" 5/13 Let tan a = sinβ‘π/cosβ‘π = (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 β cos2 b sin b = β("1 β cos2 b " ) = β("1 β " (4/5)^2 ) = β(1 β16/25) = β((25 β 16)/25) = β(9/25) = 3/5 Let tan b = sinβ‘π/cosβ‘π = (3/5)/(4/5) = 3/5 Γ 5/4 = 3/4 tan b = 3/4 We know that tan (a + b) = πππβ‘γπ +γ πππγβ‘γπ γ γ/(π β πππβ‘γπ πππβ‘π γ ) Putting tan a = 12 and tan b = 3/4 = (12/5 + 3/4)/(1 β 12/5 Γ 3/4) = ((48 +15 )/20)/((20 β 36)/20) = (63/20)/((β16)/20) = 63/20 Γ 20/(β16) = β63/( 16) Hence, tan (a + b) = β 63/16 a + b = tan-1 (β( 63)/16) Putting value of a & b sin-1 12/13 + cos-1 4/5 = tan-1 (β( 63)/16) Taking L.H.S sin-1 12/13 + cos-1 4/5 + tan-1 63/16 Putting values = tan-1 (β63/16) + tan-1 (63/16) Using tan-1x + tan-1y = tan-1((π + π)/(π β ππ)) Putting x = (β 63)/16 and y by = 63/16 = tan-1(((β 63)/16 + 63/16)/(1 β (β 63)/16 Γ 63/16)) = tan-1(0/(1 β (( 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 tanβ1 & then use tan-1 a + tan-1b formula Let a = sin-1 12/13 sin a = 12/13 We know that cos2 a = 1 β sin2 a cos a = β(1 βsin2β‘π ) =β(1β(12/13)^2 ) " =" β(1β144/169) " =" β((169 β144)/169) " =" β(25/169) " =" β(52/132) "=" 5/13 Let tan a = sinβ‘π/cosβ‘π = (12/13)/(5/13) = 12/13 Γ 13/5 = 12/5 tan a = 12/5 a = tan-1 ππ/π Let b = cos-1 4/5 cos b = 4/5 We know that sin2 b = 1 β cos2 b sin b = β("1 β cos2 b " ) = β("1 β " (4/5)^2 ) = β(1 β16/25) = β((25 β 16)/25) = β(9/25) = 3/5 Let tan b = sinβ‘π/cosβ‘π = (3/5)/(4/5) = 3/5 Γ 5/4 = = 3/4 tan b = 3/4 b = tan-1 π/π 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((π + π)/(π β ππ)) Putting x = 12/5 and y = 3/4 = tan-1 ((12/5 + 3/4)/(1 β 12/5 Γ 3/4))+ tan-1 63/16 = tan-1 (((12 Γ 4 + 3 Γ 5 )/20)/(1 β 36/20))+ tan-1 63/16 = tan-1 (((48 +15 )/20)/((20 β 36)/20))+ tan-1 63/16 = tan-1 ((63/20)/((β16)/20))+ tan-1 63/16 = tan-1 (63/20Γ20/(β16))+ tan-1 63/16 = tan-1 (β 63/( 16)) + tan-1 63/16 Using tan-1x + tan-1y = tan-1((π + π)/(π β ππ)) Putting x = (β 63)/16 and y by = 63/16 = tan-1(((β 63)/16 + 63/16)/(1 β (β 63)/16 Γ 63/16)) = tan-1(((β 63)/16 + 63/16)/(1 β (β 63)/16 Γ 63/16)) = tan-1(0/(1 β (( 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

Chapter 2 Class 12 Inverse Trigonometric Functions

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.