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# Properties of Inverse Trigonometric Functions

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#### Properties of Inverse Trigonometric Functions - Lesson Summary

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

Þ 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 Þ 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