Notes On Properties of Inverse Trigonometric Functions - Gujarat board Class 11 Maths
Property 1: (i) sin-1 (1/x) = cosec-1x , x â‰¥ 1 or x â‰¤ -1 (ii) cos-1 (1/x) = sec-1x , x â‰¥ 1 or x â‰¤ -1 (iii) tan-1 (1/x) = cot-1x , x > 0 (i) sin-1 (1/x) = cosec-1x , x â‰¥ 1 or x â‰¤ -1, Proof: Let cosec-1x = y i.e., x = cosec y â‡’ 1/x = sin y â‡’ sin-1(1/x) = y â‡’ sin-1(1/x) = cosec-1x Hence, (i) sin-1 (1/x) = cosec-1x , x â‰¥ 1 or x â‰¤ -1. (ii) cos-1 (1/x) = sec-1x , x â‰¥ 1 or x â‰¤ -1 Proof: Let sec-1x = y, ie, x = sec y â‡’ 1/x = cos y â‡’ cos-1(1/x) = y â‡’ cos-1(1/x) = sec-1x Hence cos-1 (1/x) = sec-1x , x â‰¥ 1 or x â‰¤ -1. (iii) tan-1 (1/x) = cot-1x , x > 0 Let cot-1x = y â‡’ x = cot y â‡’ 1/x = tan y â‡’ tan-1(1/x) = y â‡’ tan-1(1/x) = cot-1x â‡’ tan-1(1/x) = Cot-1 x  (Since cot-1x = y) Property 2: (i) sin-1(-x) = - sin-1(x),    x âˆˆ [-1,1] (ii) tan-1(-x) = -tan-1(x),   x âˆˆ R (iii) cosec-1(-x) = -cosec-1(x), |x| â‰¥ 1 (i) sin-1(-x) = - sin-1(x),    x âˆˆ [-1,1] Proof: Let sin-1(-x) = y â‡’ -x = sin y â‡’ x = - sin y or x = sin (-y) sin-1(x) = sin-1(sin (-y)) â‡’ sin-1x = y â‡’ sin-1x = -sin-1(-x) ['.' y = sin-1(-x)] Hence, sin-1(-x) = -sin-1x, x âˆˆ [-1,1] ii) Consider tan-1(-x) = -tan-1(x),   x âˆˆ R Let tan-1 (-x) = y tan y = -x â‡’ x = -tan y or x = tan (-y) â‡’ tan-1x = tan-1{tan(-y)} â‡’ tan-1x = -y [tan-1(tanq)=q] â‡’ tan-1x = - tan-1(-x) â‡’ tan-1 (-x) = - tan-1x (iii) cosec-1(-x) = -cosec-1(x), |x| â‰¥ 1 Proof: Let cosec-1(-x) = y â‡’ -x = cosec y â‡’ x = - cosec y or     x = cosec(-y) â‡’ cosec-1x = cosec-1(cosec(-y)) â‡’ cosec-1x = -y â‡’ cosec-1 x = - cosec-1(-x)  ['.' y = cosec-1(-x)] Hence,  cosec-1(-x) = - cosec-1 x, |x| â‰¥ 1 Property 3: (i) cos-1(-x) = Ï€ - cos-1 x, x âˆˆ [-1,1] (ii) sec-1(-x) = Ï€ - sec-1x, |x| â‰¥ 1 (iii) cot-1(-x) = Ï€ - cot-1x, x âˆˆ R (i) cos-1(-x) = Ï€ - cos-1 x, x âˆˆ [-1,1] Let cos-1(-x) = y â‡’ cos y = -x â‡’ x = -cos y â‡’ x = cos (p - y) {'.' cos (p -q) = -cosq} â‡’ cos-1 x = p - y â‡’ cos-1 x = p - Cos-1(-x) cos-1(-x) = p - cos-1 x (ii) sec-1(-x) = Ï€ - sec-1x, |x| â‰¥ 1 Proof: Let sec-1(-x) = y â‡’ -x = sec y â‡’ -x = sec y  = sec(Ï€ - y) â‡’ sec-1x  = sec-1[sec(Ï€ - y)] â‡’ sec-1x  = Ï€ - y â‡’ y  = Ï€ - sec-1x Hence, sec-1(-x) = Ï€ - sec-1x, |x| â‰¥ 1 (iii) cot-1(-x) = Ï€ - cot-1x, x âˆˆ R Proof: Let cot-1(-x) = y â‡’ -x = cot y â‡’ x = - cot y = cot(Ï€-y) â‡’ cot-1x = cot-1[cot(Ï€-y)] â‡’ cot-1x = Ï€-y â‡’ y = Ï€ - cot-1x Property 4: (i) sin-1x + cos-1x = Ï€/2, x âˆˆ [-1,1] (ii) tan-1x + cot-1x = Ï€/2, x âˆˆ R (iii) cosec-1x + sec-1x = Ï€/2, |x| â‰¥ 1 (i) sin-1x + cos-1x = Ï€/2, x âˆˆ [-1,1] Proof: Let sin-1x = y â‡’ x = sin y = cos(Ï€/2 - y) â‡’ cos-1x = cos-1[cos(Ï€/2 - y)] â‡’ cos-1x = Ï€/2 - y â‡’ cos-1x = Ï€/2 - sin-1x â‡’ sin-1x + cos-1x = Ï€/2 Hence, sin-1x + cos-1x = Ï€/2, x âˆˆ [-1,1] (ii) tan-1x + cot-1x = Ï€/2, x âˆˆ R Let tan-1 x = y â‡’ x = tan y â‡’ x = Cot (Ï€/2 - y) ['.' tan Î¸ = cot(Ï€/2 - Î¸)] â‡’ cot-1x = cot-1[Cot (Ï€/2 - y)] â‡’ cot-1x = Ï€/2 - y â‡’ cot-1 x + y = Ï€/2 â‡’ cot-1 x + tan-1x = Ï€/2 (iii) cosec-1x + sec-1x = Ï€/2, |x| â‰¥ 1   Proof: Let cosec-1x = y â‡’ x = cosec y = sec(Ï€/2 - y) â‡’ sec-1x = sec-1[sec(Ï€/2 - y)] â‡’ sec-1x = Ï€/2 - y â‡’ sec-1x = Ï€/2 - cosec-1x cosec-1x + sec-1x = Ï€/2 Hence, cosec-1x + sec-1x = Ï€/2, |x| â‰¥ 1 Property 5: (i) tan-1x + tan-1y = tan-1((x+y)/(1-xy)), xy < 1. (ii) tan-1x - tan-1y = tan-1((x-y)/(1+xy)), xy > -1. (i) tan-1x + tan-1y = tan-1((x+y)/(1-xy)), xy < 1. Let tan-1x = A and tan-1y = B Ãž tan A = x and tan B = y Now, tan(A+B) = (tanA + tanB)/(1-tanAtanB) â‡’ tan(A+B) = (x+y)/(1-xy) â‡’tan-1[(x+y)/(1-xy)] = A + B â‡’tan-1[(x+y)/(1-xy)] = tan-1x + tan-1y (ii) tan-1x - tan-1y = tan-1((x-y)/(1+xy)), xy > -1. Proof: Let tan-1x = Î± â‡’ x = tan Î±      tan-1y = Î² â‡’ y = tan Î² Now, tan(Î± - Î²) = (tan Î± - tan Î²)/(1+ tanÎ±tanÎ²) = (x-y)/(1+xy) Î± - Î² = tan-1 [(x-y)/(1+xy)] tan-1x - tan-1y = tan-1 [(x-y)/(1+xy)] Hence, tan-1x - tan-1y = tan-1 [(x-y)/(1+xy)], xy > -1 Property 6: (i) 2tan-1x = sin-1 (2x/(1+x2)), |x| â‰¤ 1 (ii) 2tan-1x = cos-1((1-x2)/(1+x2)), x â‰¥ 0 (iii) 2tan-1x = tan-1(2x/(1 - x2)), -1 < x <1 (i) 2tan-1x = sin-1 (2x/(1+x2)), |x| â‰¤ 1 Let tan-1x = y Ãž x = Tan y Consider RHS. sin-1(2x/(1+x2)) = sin-1(2tany /(1+tan2y)) = sin-1(sin2y)  ['.' 2tanÎ¸/(1+tan2Î¸)] = 2y = 2 tan-1x = LHS 2 tan-1x = sin-1 (2x/(1+x2)), |x| â‰¤ 1 (ii) 2tan-1x = cos-1((1-x2)/(1+x2)), x â‰¥ 0 Proof: Let tan-1x = y â‡’ x = tan y Now, cos-1((1-x2)/(1+x2)) = cos-1((1-tan2y)/(1+tan2y))    = cos-1(cos2y)  ('.' cos2y = ((1-tan2y)/(1+tan2y))    = 2y    = 2 tan-1x (iii) 2tan-1x = tan-1(2x/(1 - x2)), -1 < x <1 Proof: Let tan-1x = y â‡’ x = tan y Now, tan-1(2x/(1-x2)) = tan-1 (2tany/(1-tan2y))                               = tan-1(tan 2y)                               = 2y                               = 2 tan-1x 2tan-1x = tan-1(2x/(1 - x2)) Hence, 2tan-1x = tan-1(2x/(1 - x2)) , -1< x < 1

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Property 1: (i) sin-1 (1/x) = cosec-1x , x â‰¥ 1 or x â‰¤ -1 (ii) cos-1 (1/x) = sec-1x , x â‰¥ 1 or x â‰¤ -1 (iii) tan-1 (1/x) = cot-1x , x > 0 (i) sin-1 (1/x) = cosec-1x , x â‰¥ 1 or x â‰¤ -1, Proof: Let cosec-1x = y i.e., x = cosec y â‡’ 1/x = sin y â‡’ sin-1(1/x) = y â‡’ sin-1(1/x) = cosec-1x Hence, (i) sin-1 (1/x) = cosec-1x , x â‰¥ 1 or x â‰¤ -1. (ii) cos-1 (1/x) = sec-1x , x â‰¥ 1 or x â‰¤ -1 Proof: Let sec-1x = y, ie, x = sec y â‡’ 1/x = cos y â‡’ cos-1(1/x) = y â‡’ cos-1(1/x) = sec-1x Hence cos-1 (1/x) = sec-1x , x â‰¥ 1 or x â‰¤ -1. (iii) tan-1 (1/x) = cot-1x , x > 0 Let cot-1x = y â‡’ x = cot y â‡’ 1/x = tan y â‡’ tan-1(1/x) = y â‡’ tan-1(1/x) = cot-1x â‡’ tan-1(1/x) = Cot-1 x  (Since cot-1x = y) Property 2: (i) sin-1(-x) = - sin-1(x),    x âˆˆ [-1,1] (ii) tan-1(-x) = -tan-1(x),   x âˆˆ R (iii) cosec-1(-x) = -cosec-1(x), |x| â‰¥ 1 (i) sin-1(-x) = - sin-1(x),    x âˆˆ [-1,1] Proof: Let sin-1(-x) = y â‡’ -x = sin y â‡’ x = - sin y or x = sin (-y) sin-1(x) = sin-1(sin (-y)) â‡’ sin-1x = y â‡’ sin-1x = -sin-1(-x) ['.' y = sin-1(-x)] Hence, sin-1(-x) = -sin-1x, x âˆˆ [-1,1] ii) Consider tan-1(-x) = -tan-1(x),   x âˆˆ R Let tan-1 (-x) = y tan y = -x â‡’ x = -tan y or x = tan (-y) â‡’ tan-1x = tan-1{tan(-y)} â‡’ tan-1x = -y [tan-1(tanq)=q] â‡’ tan-1x = - tan-1(-x) â‡’ tan-1 (-x) = - tan-1x (iii) cosec-1(-x) = -cosec-1(x), |x| â‰¥ 1 Proof: Let cosec-1(-x) = y â‡’ -x = cosec y â‡’ x = - cosec y or     x = cosec(-y) â‡’ cosec-1x = cosec-1(cosec(-y)) â‡’ cosec-1x = -y â‡’ cosec-1 x = - cosec-1(-x)  ['.' y = cosec-1(-x)] Hence,  cosec-1(-x) = - cosec-1 x, |x| â‰¥ 1 Property 3: (i) cos-1(-x) = Ï€ - cos-1 x, x âˆˆ [-1,1] (ii) sec-1(-x) = Ï€ - sec-1x, |x| â‰¥ 1 (iii) cot-1(-x) = Ï€ - cot-1x, x âˆˆ R (i) cos-1(-x) = Ï€ - cos-1 x, x âˆˆ [-1,1] Let cos-1(-x) = y â‡’ cos y = -x â‡’ x = -cos y â‡’ x = cos (p - y) {'.' cos (p -q) = -cosq} â‡’ cos-1 x = p - y â‡’ cos-1 x = p - Cos-1(-x) cos-1(-x) = p - cos-1 x (ii) sec-1(-x) = Ï€ - sec-1x, |x| â‰¥ 1 Proof: Let sec-1(-x) = y â‡’ -x = sec y â‡’ -x = sec y  = sec(Ï€ - y) â‡’ sec-1x  = sec-1[sec(Ï€ - y)] â‡’ sec-1x  = Ï€ - y â‡’ y  = Ï€ - sec-1x Hence, sec-1(-x) = Ï€ - sec-1x, |x| â‰¥ 1 (iii) cot-1(-x) = Ï€ - cot-1x, x âˆˆ R Proof: Let cot-1(-x) = y â‡’ -x = cot y â‡’ x = - cot y = cot(Ï€-y) â‡’ cot-1x = cot-1[cot(Ï€-y)] â‡’ cot-1x = Ï€-y â‡’ y = Ï€ - cot-1x Property 4: (i) sin-1x + cos-1x = Ï€/2, x âˆˆ [-1,1] (ii) tan-1x + cot-1x = Ï€/2, x âˆˆ R (iii) cosec-1x + sec-1x = Ï€/2, |x| â‰¥ 1 (i) sin-1x + cos-1x = Ï€/2, x âˆˆ [-1,1] Proof: Let sin-1x = y â‡’ x = sin y = cos(Ï€/2 - y) â‡’ cos-1x = cos-1[cos(Ï€/2 - y)] â‡’ cos-1x = Ï€/2 - y â‡’ cos-1x = Ï€/2 - sin-1x â‡’ sin-1x + cos-1x = Ï€/2 Hence, sin-1x + cos-1x = Ï€/2, x âˆˆ [-1,1] (ii) tan-1x + cot-1x = Ï€/2, x âˆˆ R Let tan-1 x = y â‡’ x = tan y â‡’ x = Cot (Ï€/2 - y) ['.' tan Î¸ = cot(Ï€/2 - Î¸)] â‡’ cot-1x = cot-1[Cot (Ï€/2 - y)] â‡’ cot-1x = Ï€/2 - y â‡’ cot-1 x + y = Ï€/2 â‡’ cot-1 x + tan-1x = Ï€/2 (iii) cosec-1x + sec-1x = Ï€/2, |x| â‰¥ 1   Proof: Let cosec-1x = y â‡’ x = cosec y = sec(Ï€/2 - y) â‡’ sec-1x = sec-1[sec(Ï€/2 - y)] â‡’ sec-1x = Ï€/2 - y â‡’ sec-1x = Ï€/2 - cosec-1x cosec-1x + sec-1x = Ï€/2 Hence, cosec-1x + sec-1x = Ï€/2, |x| â‰¥ 1 Property 5: (i) tan-1x + tan-1y = tan-1((x+y)/(1-xy)), xy < 1. (ii) tan-1x - tan-1y = tan-1((x-y)/(1+xy)), xy > -1. (i) tan-1x + tan-1y = tan-1((x+y)/(1-xy)), xy < 1. Let tan-1x = A and tan-1y = B Ãž tan A = x and tan B = y Now, tan(A+B) = (tanA + tanB)/(1-tanAtanB) â‡’ tan(A+B) = (x+y)/(1-xy) â‡’tan-1[(x+y)/(1-xy)] = A + B â‡’tan-1[(x+y)/(1-xy)] = tan-1x + tan-1y (ii) tan-1x - tan-1y = tan-1((x-y)/(1+xy)), xy > -1. Proof: Let tan-1x = Î± â‡’ x = tan Î±      tan-1y = Î² â‡’ y = tan Î² Now, tan(Î± - Î²) = (tan Î± - tan Î²)/(1+ tanÎ±tanÎ²) = (x-y)/(1+xy) Î± - Î² = tan-1 [(x-y)/(1+xy)] tan-1x - tan-1y = tan-1 [(x-y)/(1+xy)] Hence, tan-1x - tan-1y = tan-1 [(x-y)/(1+xy)], xy > -1 Property 6: (i) 2tan-1x = sin-1 (2x/(1+x2)), |x| â‰¤ 1 (ii) 2tan-1x = cos-1((1-x2)/(1+x2)), x â‰¥ 0 (iii) 2tan-1x = tan-1(2x/(1 - x2)), -1 < x <1 (i) 2tan-1x = sin-1 (2x/(1+x2)), |x| â‰¤ 1 Let tan-1x = y Ãž x = Tan y Consider RHS. sin-1(2x/(1+x2)) = sin-1(2tany /(1+tan2y)) = sin-1(sin2y)  ['.' 2tanÎ¸/(1+tan2Î¸)] = 2y = 2 tan-1x = LHS 2 tan-1x = sin-1 (2x/(1+x2)), |x| â‰¤ 1 (ii) 2tan-1x = cos-1((1-x2)/(1+x2)), x â‰¥ 0 Proof: Let tan-1x = y â‡’ x = tan y Now, cos-1((1-x2)/(1+x2)) = cos-1((1-tan2y)/(1+tan2y))    = cos-1(cos2y)  ('.' cos2y = ((1-tan2y)/(1+tan2y))    = 2y    = 2 tan-1x (iii) 2tan-1x = tan-1(2x/(1 - x2)), -1 < x <1 Proof: Let tan-1x = y â‡’ x = tan y Now, tan-1(2x/(1-x2)) = tan-1 (2tany/(1-tan2y))                               = tan-1(tan 2y)                               = 2y                               = 2 tan-1x 2tan-1x = tan-1(2x/(1 - x2)) Hence, 2tan-1x = tan-1(2x/(1 - x2)) , -1< x < 1

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