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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 ≥ 1 or x ≤ -1

(ii) cos-1 (1/x) = sec-1x , x ≥ 1 or x ≤ -1

(iii) tan-1 (1/x) = cot-1x , x > 0

(i) sin-1 (1/x) = cosec-1x , x ≥ 1 or x ≤ -1,

Proof: Let cosec-1x = y i.e., x = cosec y

⇒ 1/x = sin y

⇒ sin-1(1/x) = y

⇒ sin-1(1/x) = cosec-1x

Hence, (i) sin-1 (1/x) = cosec-1x , x ≥ 1 or x ≤ -1.

(ii) cos-1 (1/x) = sec-1x , x ≥ 1 or x ≤ -1

Proof: Let sec-1x = y, ie, x = sec y

⇒ 1/x = cos y

⇒ cos-1(1/x) = y

⇒ cos-1(1/x) = sec-1x

Hence cos-1 (1/x) = sec-1x , x ≥ 1 or x ≤ -1.


(iii) tan-1 (1/x) = cot-1x , x > 0

Let cot-1x = y

⇒ x = cot y

⇒ 1/x = tan y

⇒ tan-1(1/x) = y

⇒ tan-1(1/x) = cot-1x ⇒ tan-1(1/x) = Cot-1 x  (Since cot-1x = y)

Property 2:

(i) sin-1(-x) = - sin-1(x),    x ∈ [-1,1]

(ii) tan-1(-x) = -tan-1(x),   x ∈ R

(iii) cosec-1(-x) = -cosec-1(x), |x| ≥ 1

(i) sin-1(-x) = - sin-1(x),    x ∈ [-1,1]

Proof: Let sin-1(-x) = y ⇒ -x = sin y

⇒ x = - sin y or

x = sin (-y)

sin-1(x) = sin-1(sin (-y))

⇒ sin-1x = y


⇒ sin-1x = -sin-1(-x) ['.' y = sin-1(-x)]

Hence, sin-1(-x) = -sin-1x, x ∈ [-1,1]

ii) Consider tan-1(-x) = -tan-1(x),   x ∈ R

Let tan-1 (-x) = y

tan y = -x

⇒ x = -tan y or x = tan (-y)

⇒ tan-1x = tan-1{tan(-y)}

⇒ tan-1x = -y [tan-1(tanq)=q]

⇒ tan-1x = - tan-1(-x)

⇒ tan-1 (-x) = - tan-1x

(iii) cosec-1(-x) = -cosec-1(x), |x| ≥ 1

Proof:

Let cosec-1(-x) = y

⇒ -x = cosec y

⇒ x = - cosec y or

    x = cosec(-y)

⇒ cosec-1x = cosec-1(cosec(-y))

⇒ cosec-1x = -y

⇒ cosec-1 x = - cosec-1(-x)  ['.' y = cosec-1(-x)]

Hence,  cosec-1(-x) = - cosec-1 x, |x| ≥ 1


Property 3:

(i) cos-1(-x) = π - cos-1 x, x ∈ [-1,1]

(ii) sec-1(-x) = π - sec-1x, |x| ≥ 1

(iii) cot-1(-x) = π - cot-1x, x ∈ R



(i) cos-1(-x) = π - cos-1 x, x ∈ [-1,1]

Let cos-1(-x) = y

⇒ cos y = -x ⇒ x = -cos y

⇒ x = cos (p - y) {'.' cos (p -q) = -cosq}

⇒ cos-1 x = p - y

⇒ cos-1 x = p - Cos-1(-x)

cos-1(-x) = p - cos-1 x

(ii) sec-1(-x) = π - sec-1x, |x| ≥ 1

Proof: Let sec-1(-x) = y

⇒ -x = sec y

⇒ -x = sec y  = sec(π - y)

⇒ sec-1x  = sec-1[sec(π - y)]

⇒ sec-1x  = π - y

⇒ y  = π - sec-1x

Hence, sec-1(-x) = π - sec-1x, |x| ≥ 1

(iii) cot-1(-x) = π - cot-1x, x ∈ R

Proof: Let cot-1(-x) = y

⇒ -x = cot y

⇒ x = - cot y = cot(π-y)

⇒ cot-1x = cot-1[cot(π-y)]

⇒ cot-1x = π-y

⇒ y = π - cot-1x


Property 4:

(i) sin-1x + cos-1x = π/2, x ∈ [-1,1]

(ii) tan-1x + cot-1x = π/2, x ∈ R

(iii) cosec-1x + sec-1x = π/2, |x| ≥ 1

(i) sin-1x + cos-1x = π/2, x ∈ [-1,1]

Proof: Let sin-1x = y

⇒ x = sin y = cos(π/2 - y)

⇒ cos-1x = cos-1[cos(π/2 - y)]

⇒ cos-1x = π/2 - y

⇒ cos-1x = π/2 - sin-1x

⇒ sin-1x + cos-1x = π/2

Hence, sin-1x + cos-1x = π/2, x ∈ [-1,1]

(ii) tan-1x + cot-1x = π/2, x ∈ R

Let tan-1 x = y

⇒ x = tan y

⇒ x = Cot (π/2 - y) ['.' tan θ = cot(π/2 - θ)]

⇒ cot-1x = cot-1[Cot (π/2 - y)]

⇒ cot-1x = π/2 - y

⇒ cot-1 x + y = π/2

⇒ cot-1 x + tan-1x = π/2

(iii) cosec-1x + sec-1x = π/2, |x| ≥ 1

 

Proof: Let cosec-1x = y

⇒ x = cosec y = sec(π/2 - y)

⇒ sec-1x = sec-1[sec(π/2 - y)]

⇒ sec-1x = π/2 - y

⇒ sec-1x = π/2 - cosec-1x

cosec-1x + sec-1x = π/2

Hence, cosec-1x + sec-1x = π/2, |x| ≥ 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)

⇒ tan(A+B) = (x+y)/(1-xy)

⇒tan-1[(x+y)/(1-xy)] = A + B

⇒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 = α ⇒ x = tan α

     tan-1y = β ⇒ y = tan β

Now, tan(α - β) = (tan α - tan β)/(1+ tanαtanβ) = (x-y)/(1+xy)

α - β = 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| ≤ 1

(ii) 2tan-1x = cos-1((1-x2)/(1+x2)), x ≥ 0

(iii) 2tan-1x = tan-1(2x/(1 - x2)), -1 < x <1


(i) 2tan-1x = sin-1 (2x/(1+x2)), |x| ≤ 1

Let tan-1x = y Þ x = Tan y

Consider RHS.

sin-1(2x/(1+x2))

= sin-1(2tany /(1+tan2y))

= sin-1(sin2y)  ['.' 2tanθ/(1+tan2θ)]

= 2y

= 2 tan-1x = LHS

2 tan-1x = sin-1 (2x/(1+x2)), |x| ≤ 1

(ii) 2tan-1x = cos-1((1-x2)/(1+x2)), x ≥ 0

Proof:

Let tan-1x = y ⇒ 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 ⇒ 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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