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# Inverse Trigonometric Identities

• Last Updated : 16 Jun, 2021

In mathematics, inverse trigonometric functions are also known as arcus functions or anti-trigonometric functions. The inverse trigonometric functions are the inverse functions of basic trigonometric functions, i.e., sine, cosine, tangent, cosecant, secant, and cotangent. It is used to find the angles with any trigonometric ratio. Inverse trigonometric functions are generally used in fields like geometry, engineering, etc. The representation of inverse trigonometric functions are:

If a = f(b), then the inverse function is

b = f-1(a)

Examples of inverse inverse trigonometric functions are sin-1x, cos-1x, tan-1x, etc.

The following table shows some trigonometric functions with their domain and range.

### Properties of Inverse Trigonometric Functions

The following are the properties of inverse trigonometric functions:

Property 1:

1. sin-1 (1/x) = cosec-1 x, for x ≥ 1 or x ≤ -1
2. cos-1 (1/x) = sec-1 x, for x ≥ 1 or x ≤ -1
3. tan-1 (1/x) = cot-1 x, for x > 0

Property 2:

1. sin-1 (-x) = -sin-1 x, for x ∈ [-1 , 1]
2. tan-1 (-x) = -tan-1 x, for x ∈ R
3. cosec-1 (-x) = -cosec-1 x, for |x| ≥ 1

Property 3

1. cos-1 (-x) = π – cos-1 x, for x ∈ [-1 , 1]
2. sec-1 (-x) = π – sec-1 x, for |x| ≥ 1
3. cot-1 (-x) = π – cot-1 x, for x ∈ R

Property 4

1. sin-1 x + cos-1 x = π/2, for x ∈ [-1,1]
2. tan-1 x + cot-1 x = π/2, for x ∈ R
3. cosec-1 x + sec-1 x = π/2 , for |x| ≥ 1

Property 5

1. tan-1 x + tan-1 y = tan-1 ( x + y )/(1 – xy), for xy < 1
2. tan-1 x – tan-1 y = tan-1 (x – y)/(1 + xy), for xy > -1
3. tan-1 x + tan-1 y = π + tan-1 (x + y)/(1 – xy), for xy >1 ; x, y >0

Property 6

1. 2tan-1 x = sin-1 (2x)/(1 + x2), for |x| ≤ 1
2. 2tan-1 x = cos-1 (1 – x2)/(1 + x2), for x ≥ 0
3. 2tan-1 x = tan-1 (2x)/(1 – x2), for -1 < x <1

### Identities of Inverse Trigonometric Function

The following are the identities of inverse trigonometric functions:

1. sin-1 (sin x) = x provided –π/2 ≤ x ≤ π/2
2. cos-1 (cos x) = x provided 0 ≤ x ≤ π
3. tan-1 (tan x) = x provided –π/2 < x < π/2
4. sin(sin-1 x) = x provided -1 ≤ x ≤ 1
5. cos(cos-1 x) = x provided -1 ≤ x ≤ 1
6. tan(tan-1 x) = x provided x ∈ R
7. cosec(cosec-1 x) = x provided -1 ≤ x ≤ ∞ or -∞ < x ≤ 1
8. sec(sec-1 x) = x provided 1 ≤ x ≤ ∞ or -∞ < x ≤ 1
9. cot(cot-1 x) = x provided -∞ < x < ∞
10. 2cos-1 x = cos-1 (2x2 – 1)
11. 2sin-1x = sin-1 2x√(1 – x2)
12. 3sin-1x = sin-1(3x – 4x3)
13. 3cos-1 x = cos-1 (4x3 – 3x)
14. 3tan-1x = tan-1((3x – x3/1 – 3x2))
15. sin-1x + sin-1y = sin-1{ x√(1 – y2) + y√(1 – x2)}
16. sin-1x – sin-1y = sin-1{ x√(1 – y2) – y√(1 – x2)}
17. cos-1 x + cos-1 y = cos-1 [xy – √{(1 – x2)(1 – y2)}]
18. cos-1 x – cos-1 y = cos-1 [xy + √{(1 – x2)(1 – y2)}
19. tan-1 x + tan-1 y = tan-1(x + y/1 – xy)
20. tan-1 x – tan-1 y = tan-1(x – y/1 + xy)
21. tan-1 x + tan-1 y +tan-1 z = tan-1 (x + y + z – xyz)/(1 – xy – yz – zx)

### Sample Problems

Question 1: Prove sin-1 x = sec-1 1/√(1-x2)

Solution:

Let sin-1 x = y

⇒ sin y = x , (since sin y = perpendicular/hypotenuse ⇒ cos y = √(1- perpendicular2 )/hypotenuse )

⇒ cos y = √(1 – x2), here hypotenuse = 1

⇒ sec y = 1/cos y

⇒ sec y = 1/√(1 – x2)

⇒ y = sec-1 1/√(1 – x2)

⇒ sin-1 x = sec-1 1/√(1 – x2)

Hence, proved.

Question 2: Prove tan-1 x = cosec-1 √(1 + x2)/x

Solution:

Let tan-1 x = y

⇒ tan y = x , perpendicular = x and base = 1

⇒ sin y = x/√(x2 + 1) , (since hypotenuse = √(perpendicular2 + base2 ) )

⇒ cosec y = 1/sin y

⇒ cosec y = √(x2 + 1)/x

⇒ y = cosec-1 √(x2 + 1)/x

⇒ tan-1 x = cosec-1 √(x2 + 1)/x

Hence, proved.

Question 3: Evaluate tan(cos-1 x)

Solution:

Let cos-1 x = y

⇒ cos y = x , base = x and hypotenuse = 1 therefore sin y = √(1 – x2)/1

⇒ tan y = sin y/ cos y

⇒ tan y = √(1 – x2)/x

⇒ y = tan-1 √(1 – x2)/x

⇒ cos-1 x = tan-1 √(1 – x2)/x

Therefore, tan(cos-1 x) = tan(tan-1 √(1 – x2)/x ) = √(1 – x2)/x.

Question 4: tan-1 √(sin x) + cot-1 √(sin x) = y. Find cos y.

Solution:

We know that tan-1 x + cot-1 x = /2 therefore comparing this identity with the equation given in the question we get y = π/2

Thus, cos y = cos π/2 = 0.

Question 5: tan-1 (1 – x)/(1 + x) = (1/2)tan-1 x, x > 0. Solve for x.

Solution:

tan-1 (1 – x)/(1 + x) = (1/2)tan-1 x

⇒ 2tan-1 (1 – x)/(1 + x) = tan-1 x     …(1)

We know that, 2tan-1 x = tan-1 2x/(1 – x2).

Therefore, LHS of equation (1) can be written as

tan-1 [ { 2(1 – x)/(1 + x)}/{ 1 – [(1 – x)(1 + x)]2}]

= tan-1 [ {2(1 – x)(1 + x)} / { (1 + x)2 – (1 – x)2 }]

= tan-1 [ 2(1 – x2)/(4x)]

= tan-1 (1 – x2)/(2x)

Since, LHS = RHS therefore

tan-1 (1 – x2)/(2x) = tan-1 x

⇒ (1 – x2)/2x = x

⇒ 1 – x2 = 2x2

⇒ 3x2 = 1

⇒ x = ± 1/√3

Since, x must be greater than 0 therefore x = 1/√3 is the acceptable answer.

Question 6: Prove tan-1 √x = (1/2)cos-1 (1 – x)/(1 + x)

Solution:

Let tan-1 √x = y

⇒ tan y = √x

⇒ tan2 y = x

Therefore,

RHS = (1/2)cos-1 ( 1- tan2 y)/(1 + tan2 y)

= (1/2)cos-1 (cos2 y – sin2 y)/(cos2 y + sin2 y)

= (1/2)cos-1 (cos2 y – sin2 y)

= (1/2)cos-1 (cos 2y)

= (1/2)(2y)

= y

= tan-1 √x

= LHS

Hence, proved.

Question 7: tan-1 (2x)/(1 – x2) + cot-1 (1 – x2)/(2x) = π/2, -1 < x < 1. Solve for x.

Solutions:

tan-1 (2x)/(1 – x2) + cot-1 (1 – x2)/(2x) = π/2

⇒ tan-1 (2x)/(1 – x2) + tan-1 (2x)/(1 – x2) = π/2

⇒ 2tan-1 (2x)/(1 – x2) = ∏/2

⇒ tan-1 (2x)/(1 – x2) = ∏/4

⇒ (2x)/(1 – x2) = tan ∏/4

⇒ (2x)/(1 – x2) = 1

⇒ 2x = 1 – x2

⇒ x2 + 2x -1 = 0

⇒ x = [-2 ± √(22 – 4(1)(-1))] / 2

⇒ x = [-2 ± √8] / 2

⇒ x = -1 ± √2

⇒ x = -1 + √2 or x = -1 – √2

But according to the question x ∈ (-1, 1) therefore for the given equation the solution set is x ∈ ∅.

Question 8: tan-1 1/(1 + 1.2) + tan-1 1/(1 + 2.3) + … + tan-1 1/(1 + n(n + 1)) = tan-1 x. Solve for x.

Solution:

tan-1 1/(1 + 1.2) + tan-1 1/(1 + 2.3) + … + tan-1 1/(1 + n(n + 1)) = tan-1 x

⇒ tan-1 (2 – 1)/(1 + 1.2) + tan-1 (3 – 2)/(1 + 2.3) + … + tan-1 (n + 1 – n)/(1 + n(n + 1)) = tan-1 x

⇒ (tan-1 2 – tan-1 1) + (tan-1 3 – tan-1 2) + … + (tan-1 (n + 1) – tan-1 n) = tan-1 x

⇒ tan-1 (n + 1) – tan-1 1 = tan-1 x

⇒ tan-1 n/(1 + (n + 1).1) = tan-1 x

⇒ tan-1 n/(n + 2) = tan-1 x

⇒ x = n/(n + 2)

Question 9: If 2tan-1 (sin x) = tan-1 (2sec x) then solve for x.

Solution:

2tan-1 (sin x) = tan-1 (2sec x)

⇒ tan-1 (2sin x)/(1 – sin2 x) = tan-1 (2/cos x)

⇒ (2sin x)/(1 – sin2 x) = 2/cos x

⇒ sin x/cos2 x = 1/cos x

⇒ sin x cos x = cos2 x

⇒ sin x cos x – cos2 x = 0

⇒ cos x(sin x – cos x) = 0

⇒ cos x = 0 or sin x – cos x = 0

⇒ cos x = cos π/2 or tan x = tan π/4

⇒ x = π/2 or x = π/4

But at x = π/2 the given equation does not exist hence x = π/4 is the only solution.

Question 10: Prove that cot-1 [ {√(1 + sin x) + √(1 – sin x)}/{√(1 + sin x) – √(1 – sin x)}] = x/2, x ∈ (0, π/4)

Solution:

Let x = 2y therefore

LHS = cot-1 [{√(1+sin 2y) + √(1-sin 2y)}/{√(1+sin 2y) – √(1-sin 2y)}]

= cot-1 [{√(cos2 y + sin2 y + 2sin y cos y) + √(cos2 y + sin2 y – 2sin y cos y)}/{√(cos2 y + sin2 y + 2sin y cos y) – √(cos2 y + sin2 y – 2sin y cos y)} ]

= cot-1 [{√(cos y + sin y)2 + √(cos y – sin y)2} / {√(cos y + sin y)2 – √(cos y – sin y)2}]

= cot-1 [( cos y + sin y + cos y – sin y )/(cos y + sin y – cos y + sin y)]

= cot-1 (2cos y)/(2sin y)

= cot-1 (cot y)

= y

= x/2.

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