Inverse Trigonometric Identities (Practice Questions)

Inverse Trigonometric Identities are mathematical formulas that involve inverse trigonometric functions such as sin⁻¹, cos⁻¹, and tan⁻¹, and are used to simplify or solve expressions involving angles.

Example: sin⁻¹(x) + cos⁻¹(x) = 90°

Solved Examples

Example 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.

Example 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.

Example 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.

Example 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.

Example 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.

Example 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.

Example 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 ∈ ∅.

Example 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)

Example 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.

Example 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.

Practice Question

Question 1: Solve for x in the equation sin^-1(x) + cos^-1(x) = π/2.

Question 2: Prove that tan^-1(1) + tan^-1(2) + tan^-1(3) = π.

Question 3: Evaluate cos⁡(sin^-1(0.5))

Question 4: If tan^-1(x) + tan^-1(2x) = π/4, then find x.

Answer:-

1. x ∈ [-1, 1]
2. π
3. √3/2.
4. (-3±√17)/4