Inverse Trigonometric Functions Notes for JEE by PW

Inverse trigonometric functions return principal angles for given trigonometric values. They are important in Class 12 Mathematics and JEE Main and Advanced because questions combine domains, ranges, graphs, identities, inequalities, derivatives, and series. Correctly selecting the principal value and checking domain restrictions are essential for solving these questions accurately.

PW supports JEE Mathematics preparation through concept-focused classes, live and recorded learning, practice questions, previous years’ questions (PYQs), and tests. These resources help you revise inverse trigonometric properties, apply identities, and practise different question patterns.

Inverse Functions: One-One and Onto Conditions

A function has an inverse function defined on its entire codomain when it is bijective, meaning it is both one-one and onto.

• One-one: Different inputs produce different outputs.

• Onto: Every element of the codomain is an output of the function.

Trigonometric functions are periodic, so they repeat values and are not one-one over their complete natural domains. To define their inverses, their domains are restricted to intervals where they are one-one.

The general process is:

1. Restrict the original domain so that the function becomes one-one.

2. Choose the codomain to equal the range of the restricted function.

3. Interchange the domain and range to define the inverse.

For inverse trigonometric functions, the input is a trigonometric value and the output is the selected principal angle.

Domains, Ranges, and Main Properties

The principal branches of the six inverse trigonometric functions are summarised below.

Here, ℝ denotes the set of real numbers. The notation sin⁻¹x means inverse sine, not 1/sin x.

The most important properties to remember are:

• Domain

• Principal range

• Parity

• Monotonicity

• Derivative

• Graph

Memory tip: Remember the order “Domain → Range → Parity → Monotonicity → Derivative.” Domain and range are particularly important when evaluating composite functions and inequalities.

Graph of a Function and Its Inverse

The graph of an inverse function is the reflection of the original function’s graph across the line:

y = x

During this reflection:

• Original x-coordinates become y-coordinates.

• Original y-coordinates become x-coordinates.

• The domain and range are interchanged.

For example, the graph of y = sin⁻¹x is the reflection of the restricted graph of y = sin x, where x ∈ [−π/2, π/2], across y = x.

The restriction is necessary because the complete sine graph is not one-one.

Derivatives of Basic Inverse Trigonometric Functions

The derivatives of inverse trigonometric functions can be obtained using implicit differentiation.

Derivative of sin⁻¹x

Let: y = sin⁻¹x

Then: sin y = x

Differentiating with respect to x: cos y · dy/dx = 1

Therefore: dy/dx = 1/cos y

Since y ∈ [−π/2, π/2], cos y ≥ 0. Using:

cos²y = 1 − sin²y

and sin y = x:

cos y = √(1 − x²)

Hence, d/dx (sin⁻¹x) = 1/√(1 − x²)

This derivative is valid for −1 < x < 1.

Derivative of cos⁻¹x

Let: y = cos⁻¹x

Then: cos y = x

Differentiating: −sin y · dy/dx = 1

Therefore, dy/dx = −1/sin y

Since sin y = √(1 − x²) on the principal range [0, π]:

d/dx (cos⁻¹x) = −1/√(1 − x²)

This derivative is valid for −1 < x < 1.

Derivative of tan⁻¹x

Let: y = tan⁻¹x

Then: tan y = x

Differentiating: sec²y · dy/dx = 1

Therefore, dy/dx = 1/sec²y

Using: sec²y = 1 + tan²y

and tan y = x: d/dx (tan⁻¹x) = 1/(1 + x²)

Derivatives of the Other Inverse Functions

The remaining standard derivatives are:

d/dx (cot⁻¹x) = −1/(1 + x²)

d/dx (cosec⁻¹x) = −1/(|x|√(x² − 1))

d/dx (sec⁻¹x) = 1/(|x|√(x² − 1))

The derivatives of sec⁻¹x and cosec⁻¹x are valid for |x| > 1. The derivatives of cot⁻¹x and tan⁻¹x are valid for all real x.

Quadrant Rules for Principal Values

The sign of the argument helps determine the principal value, but the selected angle must always lie within the function’s principal range.

For sin⁻¹x, cosec⁻¹x, and tan⁻¹x

Their principal values lie in ranges associated with the first and fourth quadrants for sine and tangent, with zero excluded for cosec⁻¹x.

For a positive argument, the principal angle is positive. For a negative argument, the principal angle is negative.

Examples:

sin⁻¹(−x) = −sin⁻¹x

tan⁻¹(−x) = −tan⁻¹x

cosec⁻¹(−x) = −cosec⁻¹x

These identities hold within the respective domains.

For cos⁻¹x, sec⁻¹x, and cot⁻¹x

Their principal values lie in [0, π], excluding π/2 for sec⁻¹x and using (0, π) for cot⁻¹x.

For cos⁻¹x, a negative argument gives a principal value in the second quadrant.

For example: cos⁻¹(−1/2) = π − cos⁻¹(1/2)

Since cos⁻¹(1/2) = π/3: cos⁻¹(−1/2) = 2π/3

Memory tip: Never select an angle using the sign alone. Check the principal range of the inverse function before finalising the answer.

Important Inverse Trigonometric Identities

Odd Function Identities

The functions sin⁻¹x, tan⁻¹x, and cosec⁻¹x are odd on their respective domains.

Therefore:

sin⁻¹(−x) = −sin⁻¹x

tan⁻¹(−x) = −tan⁻¹x

cosec⁻¹(−x) = −cosec⁻¹x

The functions cos⁻¹x, sec⁻¹x, and cot⁻¹x are neither even nor odd under their standard principal branches.

Complementary Identities

For x ∈ [−1, 1]: sin⁻¹x + cos⁻¹x = π/2

For every real x: tan⁻¹x + cot⁻¹x = π/2

For |x| ≥ 1: sec⁻¹x + cosec⁻¹x = π/2

These identities use the standard principal ranges defined in the properties table.

Relationship Between cot⁻¹x and tan⁻¹(1/x)

For x > 0: cot⁻¹x = tan⁻¹(1/x)

For x < 0: cot⁻¹x = π + tan⁻¹(1/x)

The additional π in the second case ensures that the result lies within the principal range (0, π) of cot⁻¹x.

Inverse Trigonometric Graph Shortcuts

Expressions such as sin⁻¹(sin x), cos⁻¹(cos x), and tan⁻¹(tan x) cannot always be simplified by cancelling the function and its inverse. The original angle must be adjusted to the principal range.

sin⁻¹(sin x)

The output of sin⁻¹(sin x) must lie in [−π/2, π/2].

For any integer k:

If x ∈ [−π/2 + 2kπ, π/2 + 2kπ], then: sin⁻¹(sin x) = x − 2kπ

If x ∈ [π/2 + 2kπ, 3π/2 + 2kπ], then: sin⁻¹(sin x) = (2k + 1)π − x

Examples:

sin⁻¹(sin 3) = π − 3

sin⁻¹(sin 5) = 5 − 2π

sin⁻¹(sin 10) = 3π − 10

Each result lies in the principal range [−π/2, π/2].

cos⁻¹(cos x)

The output of cos⁻¹(cos x) must lie in [0, π].

For any integer k:

If x ∈ [2kπ, (2k + 1)π], then: cos⁻¹(cos x) = x − 2kπ

If x ∈ [(2k − 1)π, 2kπ], then: cos⁻¹(cos x) = 2kπ − x

For example, since 6 lies between π and 2π: cos⁻¹(cos 6) = 2π − 6

tan⁻¹(tan x)

The output of tan⁻¹(tan x) must lie in (−π/2, π/2).

For any integer k, if: x ∈ (kπ − π/2, kπ + π/2)

Then: tan⁻¹(tan x) = x − kπ

For example, since 18 lies between 11π/2 and 13π/2: tan⁻¹(tan 18) = 18 − 6π

Memory tip: Do not assume sin⁻¹(sin x) = x, cos⁻¹(cos x) = x, or tan⁻¹(tan x) = x for every real x. Always reduce the angle to the appropriate principal range.

Inverse Trigonometric Inequalities

Monotonicity determines whether the inequality sign remains the same or reverses.

Standard Method

1. Find the domain of every inverse function.

2. Identify whether the function is increasing or decreasing.

3. Apply the corresponding trigonometric function or inverse-function property.

4. Solve the resulting inequality.

5. Intersect the result with the original domain.

For example, consider: tan⁻¹x > π/4

Since tan⁻¹1 = π/4: tan⁻¹x > tan⁻¹1

The function tan⁻¹x is increasing over ℝ, so: x > 1

Always check the original domain and range conditions before accepting a solution.

Telescoping Series of Inverse Tangents

A telescoping series is a sum in which intermediate terms cancel, leaving only a few terms.

A useful identity is: tan⁻¹u − tan⁻¹v = tan⁻¹[(u − v)/(1 + uv)]

This form holds directly when the principal-value conditions are satisfied; otherwise, an appropriate multiple of π may be required.

Standard Method

1. Factorise the denominator of each term.

2. Express each term as a difference of inverse tangents.

3. Write out the first few terms.

4. Cancel the intermediate terms.

5. Evaluate the remaining terms and check principal-value conditions.

Consider the series term: Tₙ = tan⁻¹[2n/((n² + n + 1)(n² − n + 1))]

Notice that: (n² + n + 1) − (n² − n + 1) = 2n

Also, 1 + (n² + n + 1)(n² − n + 1) = n⁴ + n² + 1

The inverse-tangent subtraction identity gives:

Tₙ = tan⁻¹(n² + n + 1) − tan⁻¹(n² − n + 1)

for positive integer n, with the principal-value condition satisfied.

For the first n terms: Sₙ = Σₖ₌₁ⁿ Tₖ

Expanding the sum:

Sₙ = [tan⁻¹3 − tan⁻¹1] + [tan⁻¹7 − tan⁻¹3] + ... + [tan⁻¹(n² + n + 1) − tan⁻¹(n² − n + 1)]

The intermediate terms cancel, leaving:

Sₙ = tan⁻¹(n² + n + 1) − tan⁻¹1

Since tan⁻¹1 = π/4:

Sₙ = tan⁻¹(n² + n + 1) − π/4

As n → ∞: limₙ→∞ tan⁻¹(n² + n + 1) = π/2

Therefore, the infinite sum is: S = π/2 − π/4 = π/4

Inverse Trigonometric Functions require a strong understanding of principal values, domains, ranges, identities, and graph behaviour. For JEE Main and JEE Advanced, practise simplifying composite functions, differentiating inverse trigonometric expressions, solving inequalities, and evaluating telescoping series. PW supports your JEE Mathematics preparation with concept-focused classes, live and recorded learning, practice questions, PYQs, and tests to help you revise formulas and strengthen problem-solving skills.

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