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CBSE Class 12 Maths Revision Notes Chapter 2 Inverse Trigonometric Functions

Inverse trigonometric functions are defined by restricting trigonometric functions to suitable domains where their inverses exist. For CBSE Class 12 Maths, this chapter focuses on principal value branches, domain, range, graphs and standard properties.

Inverse Trigonometric Functions explain how trigonometric functions can be reversed after restricting their domains. Since sine, cosine, tangent, cotangent, secant and cosecant are periodic functions, they are not one-one over their natural domains. To define their inverses, a principal value branch is selected.

Use these CBSE Class 12 Maths Revision Notes Chapter 2 for the 2026–27 academic year to revise domain and range, principal values, graphs, properties and common formulas. These Class 12 Mathematics Chapter 2 notes also help with exercise-based questions where the final answer must lie in the correct principal branch.

Key Takeaways

Inverse condition: A function has an inverse only when it is one-one and onto.
Principal value branch: It gives a unique output for every valid input.
Domain and range: Each inverse trigonometric function has a fixed domain and principal range.
Common mistake: sin⁻¹x means inverse sine, while (sin x)⁻¹ means 1/sin x.

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Access Class 12 Maths Chapter 2 Inverse Trigonometric Functions Notes in 30 Minutes

This chapter becomes easier when you revise it in the NCERT order. Start with inverse functions, then revise restricted domains, principal value branches and properties.

Revision Area	What to Revise
Inverse Function	Meaning and condition for inverse
Need for Restrictions	Why trigonometric functions are restricted
Principal Value Branch	Standard range for each inverse function
Domain and Range	Six inverse trigonometric functions
Graphs	Reflection in the line y = x
Properties	Reciprocal, negative, complementary and addition formulas
Principal Values	Standard angle-based values
Simplification	Formula use with domain and range checks

Inverse Trigonometric Functions Class 12 Notes: Basic Concept

An inverse function reverses the mapping of a function. If y = f(x), then x = f⁻¹(y), provided f is invertible.

A function is invertible only when it is one-one and onto. Trigonometric functions are not one-one over their natural domains because their values repeat after fixed intervals.

For example, sin x takes the same value at different angles. So, sin x cannot have a unique inverse unless its domain is restricted.

Why Trigonometric Functions Need Restricted Domains

Trigonometric functions are periodic. This means one output can correspond to many input angles.

For inverse trigonometric functions, each input must give only one output. To make this possible, a restricted interval is chosen where the trigonometric function becomes one-one and onto.

Trigonometric Function	Reason for Restriction
sin x	Same sine value occurs for many angles
cos x	Same cosine value occurs for many angles
tan x	Repeats after interval π
cot x	Repeats after interval π
sec x	Undefined at odd multiples of π/2
cosec x	Undefined at integral multiples of π

Principal Value Branch

A principal value branch is the standard interval chosen as the range of an inverse trigonometric function.

When no branch is mentioned, the principal value branch is used.

For example:

sin⁻¹(1/2) = π/6

Here, π/6 is selected because it lies in the principal value branch of sin⁻¹x, which is [-π/2, π/2].

Domain and Range of Inverse Trigonometric Functions

The domain and range table is the most important part of Maths Notes for Chapter 2 Inverse Trigonometric Functions Class 12. Most questions require students to check whether the given value lies in the correct domain and whether the answer lies in the correct range.

Function	Domain	Principal Value Range
y = sin⁻¹x	[-1, 1]	[-π/2, π/2]
y = cos⁻¹x	[-1, 1]	[0, π]
y = tan⁻¹x	R	(-π/2, π/2)
y = cot⁻¹x	R	(0, π)
y = sec⁻¹x	R - (-1, 1)	[0, π] - {π/2}
y = cosec⁻¹x	R - (-1, 1)	[-π/2, π/2] - {0}

sin⁻¹x: Domain, Range and Principal Value Branch

The inverse of sine is written as sin⁻¹x. It is also called arc sine.

For sin x to have an inverse, its domain is restricted to [-π/2, π/2]. On this interval, sine is one-one and onto.

Function	Domain	Range
y = sin⁻¹x	[-1, 1]	[-π/2, π/2]

If y = sin⁻¹x, then sin y = x.

Important points:

sin⁻¹x is increasing on [-1, 1].
sin⁻¹(-x) = -sin⁻¹x.
sin(sin⁻¹x) = x, where x ∈ [-1, 1].
sin⁻¹(sin x) = x, where x ∈ [-π/2, π/2].

Example:
sin⁻¹(-1/2) = -π/6

The answer lies in [-π/2, π/2], so it is a principal value.

cos⁻¹x: Domain, Range and Principal Value Branch

The inverse of cosine is written as cos⁻¹x. It is also called arc cosine.

For cos x to have an inverse, its domain is restricted to [0, π]. On this interval, cosine is one-one and onto.

Function	Domain	Range
y = cos⁻¹x	[-1, 1]	[0, π]

If y = cos⁻¹x, then cos y = x.

Important points:

cos⁻¹x is decreasing on [-1, 1].
cos⁻¹(-x) = π - cos⁻¹x.
cos(cos⁻¹x) = x, where x ∈ [-1, 1].
cos⁻¹(cos x) = x, where x ∈ [0, π].

Example:
cos⁻¹(-1/2) = 2π/3

The answer lies in [0, π], so it is a principal value.

tan⁻¹x: Domain, Range and Principal Value Branch

The inverse of tangent is written as tan⁻¹x. It is also called arc tangent.

The domain of tan x is restricted to (-π/2, π/2). On this interval, tangent becomes one-one and onto.

Function	Domain	Range
y = tan⁻¹x	R	(-π/2, π/2)

If y = tan⁻¹x, then tan y = x.

Important points:

tan⁻¹x is increasing on R.
tan⁻¹(-x) = -tan⁻¹x.
tan(tan⁻¹x) = x, where x ∈ R.
tan⁻¹(tan x) = x, where x ∈ (-π/2, π/2).

Example:
tan⁻¹(1) = π/4

The answer lies in (-π/2, π/2), so it is a principal value.

cot⁻¹x: Domain, Range and Principal Value Branch

The inverse of cotangent is written as cot⁻¹x. It is also called arc cotangent.

The domain of cot x is restricted to (0, π). On this interval, cotangent becomes one-one and onto.

Function	Domain	Range
y = cot⁻¹x	R	(0, π)

If y = cot⁻¹x, then cot y = x.

Important points:

cot⁻¹x is decreasing on R.
cot⁻¹(-x) = π - cot⁻¹x.
cot(cot⁻¹x) = x, where x ∈ R.
cot⁻¹(cot x) = x, where x ∈ (0, π).

Example:
cot⁻¹(1) = π/4

cot⁻¹(-1) = 3π/4 because the answer must lie in (0, π).

sec⁻¹x: Domain, Range and Principal Value Branch

The inverse of secant is written as sec⁻¹x. It is also called arc secant.

The domain of sec x is restricted to [0, π] - {π/2}. This makes secant one-one and onto over the selected interval.

Function	Domain	Range
y = sec⁻¹x	R - (-1, 1)	[0, π] - {π/2}

If y = sec⁻¹x, then sec y = x.

Important points:

sec⁻¹x is defined only when x ≤ -1 or x ≥ 1.
sec⁻¹(-x) = π - sec⁻¹x.
sec(sec⁻¹x) = x, where x ∈ R - (-1, 1).
sec⁻¹x is related to cos⁻¹(1/x).

Example:
sec⁻¹(2) = π/3

This is because sec π/3 = 2.

cosec⁻¹x: Domain, Range and Principal Value Branch

The inverse of cosecant is written as cosec⁻¹x. It is also called arc cosecant.

The domain of cosec x is restricted to [-π/2, π/2] - {0}. This makes cosecant one-one and onto over the selected interval.

Function	Domain	Range
y = cosec⁻¹x	R - (-1, 1)	[-π/2, π/2] - {0}

If y = cosec⁻¹x, then cosec y = x.

Important points:

cosec⁻¹x is defined only when x ≤ -1 or x ≥ 1.
cosec⁻¹(-x) = -cosec⁻¹x.
cosec(cosec⁻¹x) = x, where x ∈ R - (-1, 1).
cosec⁻¹x is related to sin⁻¹(1/x).

Example:
cosec⁻¹(2) = π/6

This is because cosec π/6 = 2.

Graphs of Inverse Trigonometric Functions

Graphs of inverse trigonometric functions are obtained by reflecting the graph of the original function in the line y = x.

This happens because inverse functions interchange input and output values. If (a, b) lies on the graph of a function, then (b, a) lies on the graph of its inverse.

Function	Graph Behaviour
y = sin⁻¹x	Increasing curve from -π/2 to π/2
y = cos⁻¹x	Decreasing curve from π to 0
y = tan⁻¹x	Increasing curve with horizontal behaviour near ±π/2
y = cot⁻¹x	Decreasing curve from π to 0
y = sec⁻¹x	Two branches for x ≤ -1 and x ≥ 1
y = cosec⁻¹x	Two branches for x ≤ -1 and x ≥ 1

Properties of Inverse Trigonometric Functions

Properties of inverse trigonometric functions help simplify expressions. They must be used only where the expressions are defined and the answer lies in the correct principal branch.

Reciprocal Identities of Inverse Trigonometric Functions

Reciprocal identities connect inverse functions of reciprocal trigonometric functions.

Identity	Condition
sin⁻¹(1/x) = cosec⁻¹x	x ∈ (-∞, -1] ∪ [1, ∞)
cos⁻¹(1/x) = sec⁻¹x	x ∈ (-∞, -1] ∪ [1, ∞)
tan⁻¹(1/x) = cot⁻¹x	x > 0

For x < 0, principal value adjustment is needed in tangent and cotangent relations.

Negative Argument Identities

These identities help simplify expressions with negative inputs.

Identity	Domain
sin⁻¹(-x) = -sin⁻¹x	x ∈ [-1, 1]
tan⁻¹(-x) = -tan⁻¹x	x ∈ R
cosec⁻¹(-x) = -cosec⁻¹x	x ∈ R - (-1, 1)
cos⁻¹(-x) = π - cos⁻¹x	x ∈ [-1, 1]
cot⁻¹(-x) = π - cot⁻¹x	x ∈ R
sec⁻¹(-x) = π - sec⁻¹x	x ∈ R - (-1, 1)

Memory point:
sin⁻¹x, tan⁻¹x and cosec⁻¹x behave like odd functions.
cos⁻¹x, cot⁻¹x and sec⁻¹x need π adjustment.

Complementary Identities

Complementary identities are useful in direct simplification questions.

Identity	Domain
sin⁻¹x + cos⁻¹x = π/2	x ∈ [-1, 1]
tan⁻¹x + cot⁻¹x = π/2	x ∈ R
sec⁻¹x + cosec⁻¹x = π/2	x ∈ R - (-1, 1)

Example:
sin⁻¹(1/2) + cos⁻¹(1/2) = π/2

π/6 + π/3 = π/2

Addition and Subtraction Formulas

These formulas are used to simplify expressions involving two inverse trigonometric terms.

Formula	Condition
tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 - xy))	xy < 1
tan⁻¹x - tan⁻¹y = tan⁻¹((x - y)/(1 + xy))	xy > -1
sin⁻¹x + sin⁻¹y = sin⁻¹(x√(1 - y²) + y√(1 - x²))	Expression must be defined
sin⁻¹x - sin⁻¹y = sin⁻¹(x√(1 - y²) - y√(1 - x²))	Expression must be defined
cos⁻¹x + cos⁻¹y = cos⁻¹(xy - √(1 - x²)√(1 - y²))	Expression must be defined
cos⁻¹x - cos⁻¹y = cos⁻¹(xy + √(1 - x²)√(1 - y²))	Expression must be defined

Always check whether the expression lies in the principal value range after applying a formula.

Double Angle Formulas in Inverse Trigonometry

These formulas help simplify expressions containing 2tan⁻¹x.

Formula	Condition
2tan⁻¹x = sin⁻¹(2x/(1 + x²))	-1 ≤ x ≤ 1
2tan⁻¹x = cos⁻¹((1 - x²)/(1 + x²))	x ≥ 0
2tan⁻¹x = tan⁻¹(2x/(1 - x²))	-1 < x < 1

These identities are useful in simplification and proof-based questions.

Principal Value Examples for Quick Revision

Principal value questions require two checks. First, find an angle that satisfies the trigonometric value. Then, check whether it lies in the required principal branch.

Example 1: Find sin⁻¹(1/2)

Let sin⁻¹(1/2) = y.

Then sin y = 1/2.

Since y must lie in [-π/2, π/2], the principal value is:

sin⁻¹(1/2) = π/6

Example 2: Find cos⁻¹(-1/2)

Let cos⁻¹(-1/2) = y.

Then cos y = -1/2.

Since y must lie in [0, π], the principal value is:

cos⁻¹(-1/2) = 2π/3

Example 3: Find tan⁻¹(-1)

Let tan⁻¹(-1) = y.

Then tan y = -1.

Since y must lie in (-π/2, π/2), the principal value is:

tan⁻¹(-1) = -π/4

Example 4: Find cot⁻¹(-1)

Let cot⁻¹(-1) = y.

Then cot y = -1.

Since y must lie in (0, π), the principal value is:

cot⁻¹(-1) = 3π/4

Example 5: Find sec⁻¹(2)

Let sec⁻¹(2) = y.

Then sec y = 2, so cos y = 1/2.

Since y must lie in [0, π] - {π/2}, the principal value is:

sec⁻¹(2) = π/3

Example 6: Find cosec⁻¹(-2)

Let cosec⁻¹(-2) = y.

Then cosec y = -2, so sin y = -1/2.

Since y must lie in [-π/2, π/2] - {0}, the principal value is:

cosec⁻¹(-2) = -π/6

Simplifying Inverse Trigonometric Expressions

Inverse trigonometric expressions often become easier after substitution. Choose a substitution that matches the inverse function.

Given Expression	Useful Substitution
sin⁻¹x	Put x = sin θ
cos⁻¹x	Put x = cos θ
tan⁻¹x	Put x = tan θ
sec⁻¹x	Put x = sec θ
cosec⁻¹x	Put x = cosec θ
cot⁻¹x	Put x = cot θ

Standard Simplification Results

Expression	Simplified Form
sin(sin⁻¹x)	x
cos(cos⁻¹x)	x
tan(tan⁻¹x)	x
cot(cot⁻¹x)	x
sec(sec⁻¹x)	x
cosec(cosec⁻¹x)	x

These results are valid when x belongs to the domain of the inverse function.

Remember These Points Before Solving

Inverse trigonometric functions often look simple, but mistakes happen due to branch and domain errors.

Do not confuse notation: sin⁻¹x is not the same as 1/sin x.
Check the input: sin⁻¹x and cos⁻¹x accept only values from [-1, 1].
Check the principal range: The final answer must lie in the correct branch.
Use conditions: Addition formulas for tan⁻¹x depend on xy.
Verify after squaring: Squaring can introduce extra roots.
Avoid invalid values: sec θ and tan θ are not defined at odd multiples of π/2.
Avoid invalid cosec and cot values: cosec θ and cot θ are not defined at θ = nπ.

Common Mistakes in Class 12 Mathematics Chapter 2 Notes

Mistake	Correct Approach
Writing sin⁻¹x as 1/sin x	sin⁻¹x means inverse sine
Ignoring principal value branch	Always check the answer range
Applying tan⁻¹ addition formula blindly	Check the condition on xy
Forgetting sec⁻¹x domain	x must be ≤ -1 or ≥ 1
Taking cot⁻¹(-1) as -π/4	cot⁻¹x has range (0, π), so answer is 3π/4
Writing cos⁻¹(-x) = -cos⁻¹x	Correct formula is cos⁻¹(-x) = π - cos⁻¹x
Assuming all inverse functions are increasing	cos⁻¹x and cot⁻¹x are decreasing

Quick Revision Summary for Inverse Trigonometric Functions

Use this table for final revision before solving NCERT exercises.

Function	Domain	Range	Behaviour
sin⁻¹x	[-1, 1]	[-π/2, π/2]	Increasing
cos⁻¹x	[-1, 1]	[0, π]	Decreasing
tan⁻¹x	R	(-π/2, π/2)	Increasing
cot⁻¹x	R	(0, π)	Decreasing
sec⁻¹x	R - (-1, 1)	[0, π] - {π/2}	Branch-based
cosec⁻¹x	R - (-1, 1)	[-π/2, π/2] - {0}	Branch-based

Important Terms in Inverse Trigonometric Functions

Term	Meaning
Inverse Function	A function that reverses another function
One-One Function	A function where different inputs have different outputs
Onto Function	A function whose range is equal to its co-domain
Principal Value	The value that lies in the principal branch
Principal Value Branch	The selected range used for an inverse trigonometric function
Domain	The set of values for which a function is defined
Range	The set of output values of a function
Arc Sine	Another name for sin⁻¹x
Arc Cosine	Another name for cos⁻¹x
Arc Tangent	Another name for tan⁻¹x

Useful Links for Class 12 Maths

Section	Useful Links
Syllabus	CBSE Class 12 Maths Syllabus
Revision Notes	CBSE Class 12 Maths Revision Notes
Maths Notes	CBSE Class 12 Maths Revision Notes Chapter 1
NCERT Solutions	NCERT Solutions Class 12 Maths
Sample Papers	CBSE Sample Papers for Class 12 Maths
Important Questions	Important Questions Class 12 Maths
NCERT Books	NCERT Books for Class 12 Maths
Previous Year Papers	CBSE Maths Question Paper Class 12

Q.1

$If \sin^{- 1} (\frac{3}{5}) = x, find the value of \cos x .$

Ans

$\begin{array}{l} Here, \sin^{- 1} \frac{3}{5} = x \\ \Rightarrow sinx = \frac{3}{5} \\ Now, \cos x = \sqrt{1 - \sin^{2} x} \\ = \sqrt{1 - {(\frac{3}{5})}^{2}} \\ = \sqrt{\frac{16}{25}} = \frac{4}{5} \end{array}$

Q.2

${Solvetan}^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4}$

Ans

$\begin{array}{l} \tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4} \\ \Rightarrow \tan^{- 1} [\frac{2 x + 3 x}{1 - 2 x \times 3 x}] = \frac{π}{4} [∴ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} [\frac{x + y}{1 - xy}]] \\ \Rightarrow \tan^{- 1} [\frac{5 x}{1 - 6 x^{2}}] = \frac{π}{4} \\ ∴ \frac{5 x}{1 - 6 x^{2}} = \tan \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \\ or 5 x = 1 - 6 x^{2} \\ \Rightarrow 6 x^{2} + 5 x - 1 = 0 \\ i . e . (6 x - 1) (x + 1) = 0 \\ Which gives, x = \frac{1}{6}, x = - 1 \\ Since x = -1 does not satisfy the equation as the L . H . S . of the equation \\ becomes negative, x = \frac{1}{6} is the only solution of the given equation . \end{array}$

Q.3

${Prove that tan}^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x + \sqrt{1 - x}}}) = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x$

Ans

$\begin{array}{l} Here, L . H . S . = {tam}^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) \\ Let x = \cos 2 θ \Rightarrow θ = \frac{1}{2} \cos^{- 1} x \\ Then L . H . S .= \tan^{- 1} [\frac{\sqrt{1 + \cos 2 θ} - \sqrt{1 - \cos 2 θ}}{\sqrt{1 + \cos 2 θ} + \sqrt{1 - \cos 2 θ}}] \\ = \tan^{- 1} [\frac{\sqrt{2 \cos^{2} θ} - \sqrt{2 \sin^{2} θ}}{\sqrt{2 \cos^{2} θ} + \sqrt{2 \sin^{2} θ}}] \\ = \tan^{- 1} [\frac{\sqrt{2} cosθ - \sqrt{2} sinθ}{\sqrt{2} cosθ + \sqrt{2} sinθ}] \\ = \tan^{- 1} [\frac{cosθ - sinθ}{cosθ + sinθ}] \\ = \tan^{- 1} [\frac{1 - tanθ}{1 + tanθ}] \\ = \tan^{- 1} [\frac{\tan \frac{π}{4} - tanθ}{1 + \tan \frac{π}{4} tanθ}] \\ = \tan^{- 1} [\tan^{- 1} (\frac{π}{4} - θ)] \\ = \frac{π}{4} - θ \\ = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x = R . H . S Proved \end{array}$

Q.4

${Prove that cot}^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] = \frac{x}{2}, \times \in (0, \frac{π}{4}) .$

Ans

$\begin{array}{l} L . H . S . \\ = \cot^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] \\ = \cot^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}} \times \frac{\sqrt{1 + sinx} + \sqrt{1 + sinx}}{\sqrt{1 + sinx} + \sqrt{1 - sinx}}] \\ = \cot^{- 1} [\frac{{(\sqrt{1 + sinx} + \sqrt{1 - sinx})}^{2}}{1 + sinx - 1 + sinx}] \\ = \cot^{- 1} [\frac{1 + sinx + 1 - sinx + 2 \sqrt{1 - \sin^{2} x}}{2 sinx}] \\ = \cot^{- 1} [\frac{2 (1 + \cos x)}{2 \sin x}] \\ = \cot^{- 1} [\frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}] \\ = \cot^{- 1} [\cot \frac{x}{2}] \\ = \frac{x}{2} = R . H . S \end{array}$

Q.5

${Show that sin}^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{63}{16} = π$

Ans

$\begin{array}{l} \sin^{- 1} \frac{12}{13} = x, \cos^{- 1} \frac{4}{5} = y \tan^{- 1} \frac{63}{16} = z \\ \Rightarrow \sin^{} x = \frac{12}{13}, \cos y = \frac{4}{5} and \tan z \frac{63}{16} \\ ∴ \cos x = \frac{5}{13}, \sin y = \frac{3}{4}, tanx = \frac{12}{5} and \tan y \frac{3}{4} \\ Now, \tan (x + y) = \frac{tanx + tany}{1 - \tan \times tany} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \times \frac{3}{4}} = \frac{- 63}{16} \\ \Rightarrow \tan (x + y) = - tanz \\ \Rightarrow \tan (x + y) = \tan (π - z) \\ ∴ x + y = π - z \\ \Rightarrow x + y + z = π \end{array}$

Q.6

$\begin{array}{l} Solve the following equation : \\ \tan^{- 1} (\frac{x - 1}{x - 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4} \end{array}$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{x - 1}{x - 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \times \frac{x + 1}{x + 2}}) = \frac{π}{4} \\ \Rightarrow \frac{x^{2} + x - 2 + x^{2} - x - 2}{x^{2} - 4 - x^{2} + 1} = 1 \\ \Rightarrow \frac{2 x^{2} - 4}{- 3} = 1 \\ \Rightarrow 2 x^{2} - 4 = - 3 \\ \Rightarrow x^{2} = \frac{1}{2} \\ \Rightarrow x = \pm \frac{1}{\sqrt{2}} \end{array}$

Q.7

$Evaluate \tan \frac{1}{2} [\sin^{- 1} \frac{2 x}{1 + x^{2}} + \cos^{- 1} \frac{1 - y^{2}}{1 + y^{2}}], |x| < 1, y > 0 and xy > 1$

Ans

$\begin{array}{l} Let x = \tan θ \\ \Rightarrow θ \tan^{- 1} x and = \tan^{- 1} y then, \\ \tan \frac{1}{2} [\sin^{- 1} \frac{2 x}{1 + x^{2}} + \cos^{- 1} \frac{1 - y^{2}}{1 + y^{2}}] \\ = \tan \frac{1}{2} [\sin^{- 1} (\frac{2 tanθ}{1 + \tan^{2} θ}) + \cos^{- 1} (\frac{1 - \tan^{2}}{1 + \tan^{2}})] \\ = \tan \frac{1}{2} [\sin^{- 1} (\sin 2 θ) + \cos^{- 1} (\cos 2)] \\ = \tan \frac{1}{2} (2 θ + 2) \\ = \tan (θ +) \end{array}$

Q.8

$Show that \sin^{- 1} \frac{3}{5} - \sin^{- 1} \frac{8}{17} = \cos^{- 1} \frac{84}{85}$

Ans

$\begin{array}{l} Let \sin^{- 1} \frac{3}{5} = x and \sin^{- 1} \frac{8}{17} = y \\ \Rightarrow \sin^{- 1} \frac{3}{5} = sinx and \frac{8}{17} = siny \\ as cosx = \sqrt{1 - \sin^{- 2} x} \\ = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \\ and \cos y = \sqrt{1 - \sin^{2} y} \\ = \sqrt{1 - \frac{64}{289}} = \frac{15}{17} \\ Now, \cos (x - y) = cosx cosy + \sin x siny \\ = \frac{4}{5} \times \frac{15}{17} + \frac{3}{5} \times \frac{8}{17} \\ \cos (x - y) = \frac{84}{85} \\ \Rightarrow x - y = \cos^{- 1} \frac{84}{85} \end{array}$

Q.9

$Evaluate \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2})$

Ans

$\begin{array}{l} \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2}) \\ \tan^{- 1} (\tan \frac{π}{4}) + \cos^{- 1} (- \cos \frac{π}{3}) + \sin^{- 1} (- \sin \frac{π}{6}) \\ = \frac{π}{4} + \cos^{- 1} (\cos (π - \frac{π}{3})) + \sin^{- 1} [\sin (\frac{- π}{6})] \\ we know range of \\ \tan^{- 1} \to (\frac{- π}{2}, \frac{π}{2}), \\ \cos^{- 1} \to [0, π] and \sin^{- 1} \to [\frac{- π}{2}, \frac{π}{2}] \\ ∴ \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2}) \\ = \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6} \\ = \frac{3 π + 8 π - 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4} \end{array}$

Q.10

$Write the function \tan^{- 1} (\frac{cosx - sinx}{cosx + sinx}) in the simplest form .$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{cosx - sinx}{cosx + sinx}) \\ = \tan^{- 1} (\frac{\frac{cosx}{cosx} - \frac{\sin x}{\cos x}}{\frac{cosx}{cosx} + \frac{sinx}{cosx}}) \\ = \tan^{- 1} (\frac{1 - tanx}{1 + tanx}) \\ = \tan^{- 1} (\frac{\tan \frac{π}{4} - tanx}{1 + \tan \frac{π}{4} tanx}) \\ = \tan^{- 1} (\tan (\frac{π}{4} - x)) [∴ \tan (A - B) = \frac{tanA - tanB}{1 + tanA tanB}] \\ = \frac{π}{4} - x \end{array}$

Q.11

${Express tan}^{-1} (\frac{cos x}{1-sin x}), \frac{π}{2} < x< \frac{π}{2} in the simplest form.$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{\cos x}{1 - \sin x}) = \tan^{- 1} [\frac{\sin (\frac{π}{2} - x)}{1 - \cos (\frac{π}{2} - x)}] \\ = \tan^{- 1} [\frac{\sin (\frac{π - 2 x}{2})}{1 - \cos (\frac{π - 2 x}{2})}] \\ = \tan^{- 1} [\frac{2 \sin (\frac{π - 2 x}{4})}{2 \sin^{2} (\frac{π - 2 x}{4})}] \\ = \tan^{- 1} [\frac{\cos (\frac{π - 2 x}{4})}{\sin (\frac{π - 2 x}{4})}] \\ = \tan^{- 1} [\cot (\frac{π - 2 x}{4})] \\ = \tan^{- 1} [\tan (\frac{π}{2} - [\frac{π - 2 x}{4}])] \\ = \tan^{- 1} [\tan (\frac{π}{4} + \frac{π}{2})] = \frac{π}{4} + \frac{x}{2} \end{array}$

Q.12

$Find the value of \sin^{- 1} (\sin \frac{2 π}{3})$

Ans

$\begin{array}{l} The range of inverse function is [\frac{- π}{2}, \frac{- π}{2}] . \\ Now, \sin^{- 1} [\sin (\frac{2 π}{3})] = \sin^{- 1} [\sin (π - \frac{π}{3})] \\ = \sin^{- 1} [\sin (\frac{π}{3})] \\ = \frac{π}{3} \end{array}$

Q.13

$Write \tan^{- 1} (\frac{1}{\sqrt{x^{2} - 1}}), |x| > 1 in the simplest form .$

Ans

$\begin{array}{l} Let x = cosec θ \\ \Rightarrow θ = {cosec}^{- 1} x \\ Then, \tan^{- 1} (\frac{1}{\sqrt{x^{2} - 1}}) = \tan^{- 1} (\frac{1}{\sqrt{{cosec}^{2} - 1}}) \\ = \tan^{- 1} (\frac{1}{cotθ}) \\ = \tan^{- 1} (tanθ) \\ = θ \\ = {cosec}^{- 1} x \end{array}$

Q.14

$Prove that \tan^{- 1} x + \tan^{- 1} \frac{2 x}{1 - x^{2}} = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$

Ans

$\begin{array}{l} Let x = \tan θ \\ Now, \\ L . H . S . = \tan^{- 1} x + \tan^{- 1} \frac{2 x}{1 - x^{2}} \\ = \tan^{- 1} (tanθ) + \tan^{- 1} (\frac{2 tanθ}{1 - \tan^{2} θ}) \\ = θ + \tan^{- 1} (\tan 2 θ) \\ = θ + 2 θ = 3 θ . \dots.. (1) \\ R . H . S .= \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (3 tanθ) = 3 θ . \dots. (2) \\ by (1) and (2) \\ L . H . S .= R . H . S . proved \end{array}$

Q.15

$Show that : \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{2}{11} = \tan^{- 1} \frac{3}{4}$

Ans

$\begin{array}{l} L . H . S .= \tan^{- 1} [\frac{\frac{1}{2} + \frac{2}{11}}{1 - \frac{1}{2} \times \frac{2}{11}}] [\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} \frac{15}{20} \\ = \tan^{- 1} \frac{3}{4} = R . H . S . \end{array}$

Q.16

$\begin{array}{l} Find the principal value of \\ (i) {sin}^{-1} (\frac{-1}{2}) \\ (ii) {cot}^{-1} (\frac{-1}{\sqrt{3}}) \end{array}$

Ans

$\begin{array}{l} (i) \sin^{- 1} (\frac{- 1}{2}) = y \\ \Rightarrow \sin y = \frac{- 1}{2} \\ \Rightarrow \sin y = - \sin \frac{π}{6} \\ \Rightarrow \sin y = - \sin (\frac{- π}{6}) or y = \frac{- π}{6} \\ We know that the range of inverse sine function is [\frac{- π}{2}, \frac{π}{2}] \\ ∴ \sin^{- 1} (\frac{- π}{6}) = \frac{- π}{6} \\ (ii) Let \cot^{- 1} (\frac{- 1}{\sqrt{3}}) = y \\ \Rightarrow coty = \frac{- 1}{\sqrt{3}} \\ \Rightarrow coty = \frac{- 1}{\sqrt{3}} \\ \Rightarrow coty = \cot (π - \frac{π}{3}) \\ \Rightarrow y = \frac{2 π}{3} \\ We know that range of inverse contangent function (0, π) \\ ∴ \cot^{- 1} (\frac{- 1}{\sqrt{3}}) = \frac{2 π}{3} \end{array}$

Q.17

$\begin{array}{l} Is the following statement true ? \\ \cos^{- 1} x = {(cosx)}^{- 1} \end{array}$

Ans

$No, \cos^{- 1} x ¹ \neq {(\cos x)}^{- 1} = \frac{1}{\cos x}$

Q.18

$Find the principal value of \cos^{- 1} (\frac{\sqrt{3}}{2})$

Ans

$\begin{array}{l} Let \cos^{- 1} (\frac{\sqrt{3}}{2}) = y \\ \Rightarrow \cos y = \frac{\sqrt{3}}{2} \\ Range of \cos^{- 1} is [0, π] \\ \cos \frac{π}{6} = \frac{\sqrt{3}}{2} \\ \Rightarrow \cos^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{6} \end{array}$

Q.19

$If \sin^{- 1} x = y, than what will be the range of y ?$

Ans

$\begin{array}{l} principal value of \sin^{- 1} lies between [- \frac{π}{2}, \frac{π}{2}] \\ - \frac{π}{2} \leq \frac{π}{2} y \leq \frac{π}{2} \end{array}$

Q.20

${Show that sin}^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \sin^{- 1} x$

Ans

$\begin{array}{l} \sin^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \sin^{- 1} x \\ let x = sinθ \\ \sin^{- 1} (2 x \sqrt{1 - x^{2}}) = \sin^{- 1} (2 sinθ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 sinθ cosθ) \\ = \sin^{- 1} (\sin 2 θ) \\ = 2 θ \\ = 2 \sin^{- 1} x \end{array}$

Q.21

$Find the value of \cos^{- 1} (\cos \frac{13 π}{6}) .$

Ans

$\begin{array}{l} \cos^{- 1} (\cos \frac{13 π}{6}) = \frac{13 π}{6} \\ But \frac{13 π}{6} \notin [0, π] \\ \cos (\frac{13 π}{6}) = \cos (2 π + \frac{π}{6}) = \cos (\frac{π}{6}) \\ ∴ \cos^{- 1} (\cos \frac{13 π}{6}) = \frac{π}{6} \end{array}$

Q.22

$Find the value of \tan^{- 1} (\frac{2}{11}) + \tan^{- 1} (\frac{7}{24}) .$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{2}{11}) + \tan^{- 1} (\frac{7}{24}) . \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) [\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} (\frac{125}{250}) \\ = \tan^{- 1} (\frac{1}{2}) \end{array}$

Q.23

$Find the value of \tan^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}) .$

Ans

$\begin{array}{l} Let x = a \tan θ \\ Then \tan^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}) = \tan^{- 1} (\frac{3 a^{2} tanθ - a^{3} \tan^{3} θ}{a^{3} - 3 a^{2} \tan^{2} θ}) \\ = \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (\tan 3 θ) \\ =3 θ =3 \tan^{- 1} (\frac{x}{a}) \end{array}$

Q.24

$Write 2 \tan^{- 1} x in terms of \sin^{- 1} x, \cos^{- 1} x and \tan^{- 1} x .$

Ans

$2 \tan^{- 1} x = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$

Q.25

Write the domain and range of tan^-1x.

Ans

$\begin{array}{l} Domain of \tan^{- 1} x = R \\ Range of {ten}^{- 1} x = (- \frac{π}{2}, \frac{π}{2}) \end{array}$

Q.26

Fill in the following blanks:

${a)cos}^{- 1} (- x) = . \dots b) \cos^{- 1} \frac{1}{x} = . \dots.$

Ans

$\begin{array}{l} a) \cos^{- 1} (- x) = π - \cos^{- 1} x \\ b) \cos^{- 1} \frac{1}{x} = \sec^{- 1} x \end{array}$

Q.27

$Find the value of sin [\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})] .$

Ans

$\begin{array}{l} \sin [\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})] \\ = \sin [\frac{π}{3} + \sin^{- 1} (- \frac{1}{2})] \\ = \sin [\frac{π}{3} + \frac{π}{6}] \\ = \sin \frac{π}{2} \\ = 1 . \end{array}$

Q.28

$Evaluate \tan^{- 1} (\tan \frac{4 π}{3}) .$

Ans

$\begin{array}{l} The rage of inverse tangent function is (\frac{- π}{2}, \frac{π}{2}) . \\ Now, \tan^{- 1} [\tan (\frac{4 π}{3})] = \tan^{- 1} [\tan (π + \frac{π}{3})] \\ = \tan^{- 1} (\frac{π}{3}) \\ = \frac{π}{3} \end{array}$

Q.29

$Solve sin \{2 \cos^{- 1} (\cot (2 \tan^{- 1} x))\} = 0 .$

Ans

$\begin{array}{l} Given equation is \\ \sin \{2 \cos^{- 1} (\cot (2 \tan^{- 1} x))\} = 0 \\ \Rightarrow 2 \cos^{- 1} (\cot (2 \tan^{- 1} x)) = n^{π}, where n is any integer \\ \Rightarrow \cos^{- 1} (\cot (2 \tan^{- 1} x)) = n^{π / 2} \\ \Rightarrow \cos^{- 1} (\cot (2 \tan^{- 1} x)) = 0, \frac{π}{2}, π \\ (since \cos^{- 1} x lies in [0, π]) \\ \Rightarrow (\cot (2 \tan^{- 1} x)) = \cos 0, \cos \frac{π}{2}, \cos π = 1, 0, - 1 \\ \Rightarrow \cot [\tan^{- 1} \{(2 x) / (1 - x^{2})\}] = 1, 0, - 1 \\ \Rightarrow 1 / [\tan \{\tan^{- 1} [(2 x) / (1 - x^{2})]\}] = 1, 0, - 1 \\ \Rightarrow (1 - x^{2}) / (2 x) = 1, 0, - 1 \\ \Rightarrow (1 - x^{2}) / (2 x) = 1, (1 - x^{2}) / (2 x) = 0 and (1 - x^{2}) / (2 x) = - 1 \\ \Rightarrow x = - 1 \pm \sqrt{2}, x = \pm 1, x = 1 \pm \sqrt{2} . \end{array}$

Q.30

${Prove that sin}^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} = \sinh - 1 \frac{63}{65} .$

Ans

$\begin{array}{l} \sin^{- 1} \frac{12}{13} = x and \cos^{- 1} \frac{4}{5} = y \\ ∴ \cos x \frac{5}{13} and siny = \frac{3}{5} \\ Now, \sin (x + y) = sinx cosy + cosx siny \\ = \frac{12}{13}, \frac{4}{5} + \frac{5}{13}, \frac{3}{5} = \frac{63}{65} \\ \Rightarrow x + y = \sin^{- 1} \frac{63}{65} \\ so, \sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} = \sin^{- 1} \frac{63}{65} \end{array}$

Q.31

$Prove that \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} = \tan^{- 1} \frac{11}{29}$

Ans

$\begin{array}{l} Prove that \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} = \tan^{- 1} \frac{11}{29} \\ L . H . S .= \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} \\ = \tan^{- 1} [\frac{\frac{1}{5} + \frac{1}{6}}{1 - \frac{1}{5} \times \frac{1}{6}}] \\ = \tan^{- 1} [\frac{6 + 5}{30 - 1}] \\ = \tan^{- 1} [\frac{11}{29}] = R . H . S . Proved . \end{array}$

Q.32

${Find the principal value of sin}^{- 1} (\frac{1}{2}) .$

Ans

$\begin{array}{l} Let \sin^{- 1} (\frac{1}{2}) = θ \\ \Rightarrow \sin θ = \frac{1}{2} \\ \Rightarrow \sin θ = \sin (\frac{π}{6}) \\ \Rightarrow \sin θ = \sin (\frac{π}{6}) (âˆµ The range of \sin -^{- 1} x is [\frac{π}{2}, \frac{π}{2}]) \\ \Rightarrow θ = \frac{π}{6} \\ ∴ The principal value of \sin^{- 1} (\frac{1}{2}) is \frac{π}{6} . \end{array}$

Q.33

${Final the principal value of sin}^{- 1} (\frac{1}{\sqrt{2}}) .$

Ans

$\begin{array}{l} Let \sin^{- 1} (\frac{1}{\sqrt{2}}) = θ \\ \Rightarrow \sin θ = \frac{1}{\sqrt{2}} \\ \Rightarrow \cos θ = - \cos (\frac{π}{4}) \\ \Rightarrow \sin θ = \sin (\frac{π}{4}) (âˆµ The range of \sin -^{- 1} x is [\frac{π}{2}, \frac{π}{2}]) \\ \Rightarrow θ = \frac{π}{4} \\ ∴ The principal value of \sin^{- 1} (\frac{1}{\sqrt{2}}) is \frac{π}{4} . \end{array}$

Q.34

$Find the principal value of \cos^{- 1} (\frac{1}{\sqrt{2}}) .$

Ans

$\begin{array}{l} Let \cos^{- 1} (- \frac{1}{\sqrt{2}}) = 0 \\ \Rightarrow \cos θ = - \frac{1}{\sqrt{2}} \\ \Rightarrow \cos θ = - \cos (\frac{π}{4}) \\ \Rightarrow \cos θ = \cos (\frac{π}{4}) (âˆµ The range of \cos -^{- 1} x is [0, π]) \\ \Rightarrow \cos θ = \cos (\frac{3 π}{4}) \\ \Rightarrow θ = - \frac{3 π}{4} \\ ∴ The principal value of \cos^{- 1} (- \frac{1}{\sqrt{2}}) is \frac{3 π}{4} . \end{array}$

Q.35

$Evaluate the \cot (\tan^{- 1} a + \cot^{- 1} a) .$

Ans

$\begin{array}{l} We have, \\ = \cot (\tan^{- 1} a + \cot^{- 1} a) \\ = \cot (\frac{π}{2}) [âˆµ \tan^{- 1} a + \cot^{- 1} a = \frac{π}{2}] \\ = 0 \end{array}$

Q.36

$Evaluate the sin ({sin}^{-1} {x+cos}^{-1} x) .$

Ans

$\begin{array}{l} We have, \\ = \sin (\sin^{- 1} x + \cos^{- 1} x) \\ = \sin (\frac{π}{2}) [âˆµ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}] \\ = 1 \end{array}$

Q.37

$Put in the simplest form : \tan^{- 1} (\frac{1 - cosθ}{sinθ}) .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{1 - \cos θ}{\sin θ}) \\ = \tan^{- 1} (\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}) \\ = \tan^{- 1} (\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}) \\ = \tan^{- 1} (\tan \frac{θ}{2}) \\ = \frac{θ}{2} \end{array}$

Q.38

${Prove that the sin}^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}} .$

Ans

$\begin{array}{l} Let \sin^{- 1} x = θ . \dots.. (1) \\ \Rightarrow \sin θ = x \\ \Rightarrow \sin^{2} θ = x^{2} \\ \Rightarrow 1 - \cos^{2} θ = x^{2} \\ \Rightarrow \cos^{2} θ = 1 - x^{2} \\ \Rightarrow \cos θ = \sqrt{1 - x^{2}} \\ \Rightarrow \cos θ = \sqrt{1 - x^{2}} \\ \Rightarrow θ = \cos^{- 1} \sqrt{1 - x^{2}} . \dots. (2) \\ From (1) and (2) we get, \\ \sin^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}} \end{array}$

Q.39

${Prove that cos}^{- 1} (- x) = π - \cos^{- 1} (x) .$

Ans

$\begin{array}{l} Let \cos^{- 1} (- x) = θ . \dots. (1) \\ \Rightarrow - x = \cos θ \\ \Rightarrow x = - \cos θ \\ \Rightarrow x = - \cos (π - θ) \\ \Rightarrow π - θ = \cos^{- 1} x \\ \Rightarrow θ = π - \cos^{- 1} x . \dots (2) \\ From (1) and (2) we get, \\ \Rightarrow \cos^{- 1} (- x) = π - \cos^{- 1} x \end{array}$

Q.40

$Evaluate the cos ({cosec}^{- 1} x + \sec^{- 1} x) .$

Ans

$\begin{array}{l} We have, \\ \cos ({cosec}^{- 1} x + \sec^{- 1} x) \\ = \cos (\frac{π}{2}) [âˆµ {cosec}^{- 1} x + \sec^{- 1} x \frac{π}{2}] \\ = 0 \end{array}$

Q.41

${Evaluate the tan}^{- 1} 2 + \tan^{- 1} 3 .$

Ans

$\begin{array}{l} {We have, tan}^{- 1} 2 + {tan}^{- 1} 3 {=tan}^{- 1} (\frac{2 + 3}{1 - 2 \times 3}) \\ = {tan}^{- 1} (\frac{5}{1 - 6}) \\ = {tan}^{- 1} (\frac{5}{- 5}) \\ = {tan}^{- 1} (- 1) \\ = {tan}^{- 1} (1) \\ = \frac{π}{4} \end{array}$

Q.42

${Evaluate the tan}^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} .$

Ans

$\begin{array}{l} We have, \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{1 - \frac{14}{11 \times 24}}) \\ = \tan^{- 1} (\frac{125}{264 - 14}) \\ = \tan^{- 1} (\frac{125}{250}) \\ = \tan^{- 1} (\frac{1}{2}) \end{array}$

Q.43

$Prove that \sin^{- 1} x = {cosec}^{- 1} \frac{1}{x} .$

Ans

$\begin{array}{l} {Letsin}^{- 1} x = θ . \dots\dots. (1) \\ \Rightarrow \sin θ = x \\ \Rightarrow cosec θ = \frac{1}{x} \\ \Rightarrow θ = {cosec}^{- 1} = (\frac{1}{x}) . \dots\dots (2) \\ From (1) and (2) we get, \\ \sin^{- 1} x = {cosec}^{- 1} (\frac{1}{x}) \end{array}$

Q.44

$Put in the simplest form, \tan^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x} .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{\sqrt{1 + x^{2} - 1}}{x}) \\ Put x = tanθ \\ = \tan^{- 1} (\frac{\sqrt{1 + \tan^{2} θ - 1}}{tanθ}) \\ = \tan^{- 1} (\frac{\sqrt{\sec^{2} θ - 1}}{tanθ}) \\ = \tan^{- 1} (\frac{secθ - 1}{tanθ}) \\ = \tan^{- 1} (\frac{\frac{1 - cosθ}{cosθ}}{\frac{sinθ}{cosθ}}) \\ = \tan^{- 1} (\frac{1 - cosθ}{sinθ}) \\ = \tan^{- 1} (\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}) \\ = \tan^{- 1} (\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}) \\ = \tan^{- 1} (\tan \frac{θ}{2}) \\ = \frac{θ}{2} [âˆµ tanθ = x \Rightarrow θ = \tan^{- 1} x] \\ = \frac{1}{2} \tan^{- 1} x \end{array}$

Q.45

${Prove that sin}^{- 1} x + \cos^{- 1} x = \frac{π}{2} .$

Ans

$\begin{array}{l} We have to prove that \\ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} . \\ Let \sin^{- 1} x = θ . \dots.. (1) \\ \Rightarrow \sin θ = x \\ \Rightarrow \cos (\frac{π}{2} - θ) = x [âˆµ sinθ = \cos (\frac{π}{2} - θ)] \\ \Rightarrow \frac{π}{2} - θ = \cos^{- 1} x \\ \Rightarrow \cos^{- 1} x = \frac{π}{2} - θ \\ \Rightarrow θ + \cos^{- 1} x = \frac{π}{2} \\ \Rightarrow \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} (using (1)) \end{array}$

Q.46

$Show that \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{1}{3} = \frac{π}{4} .$

Ans

$\begin{array}{l} We have, \\ \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{1 - \frac{14}{11 \times 24}}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{125}{264 - 14}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{125}{250}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}}) \\ = \tan^{- 1} (\frac{\frac{5}{6}}{\frac{5}{6}}) \\ = \tan^{- 1} (1) \\ = \frac{π}{4} \end{array}$

Q.47

$\begin{array}{l} If two angles of a triangle are \tan^{- 1} 2 and \tan^{- 1} 3, \\ than find the third angle . \end{array}$

Ans

$\begin{array}{l} Let third angle of a triangle be θ, then we have \\ θ + \tan^{- 1} + \tan^{- 1} 3 = π [âˆµ A + B + C = π] \\ \Rightarrow θ + \tan^{- 1} (\frac{2 + 3}{1 - 2 \times 3}) = π \\ \Rightarrow θ + \tan^{- 1} (\frac{5}{1 - 6}) = π \\ \Rightarrow θ + \tan^{- 1} (\frac{5}{- 5}) = π \\ \Rightarrow θ + \tan^{- 1} (1) = π \\ \Rightarrow θ + \frac{3 π}{4} = π [âˆµ \tan^{- 1} (- 1) = \frac{3 π}{4}] \\ \Rightarrow θ = π - \frac{3 π}{4} \\ \Rightarrow θ + \frac{π}{4} \\ Therefore, the third angle of a triangle is \frac{π}{4} . \end{array}$

Q.48

$Solve for x, \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{π}{2} .$

Ans

$\begin{array}{l} We have, \\ \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{π}{2} \\ \Rightarrow \sin^{- 1} (1 - x) = \frac{π}{2} + 2 \sin^{- 1} x \\ \Rightarrow (1 - x) = \sin (\frac{π}{2} + 2 \sin^{- 1} x) \\ \Rightarrow (1 - x) = \cos (2 \sin^{- 1} x) \\ \Rightarrow \cos^{- 1} (1 - x) = 2 \sin^{- 1} x \\ \Rightarrow \sin^{- 1} \sqrt{1 - {(1 - x)}^{2}} = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) \\ \Rightarrow \sqrt{1 - {(1 - x)}^{2}} = (2 x \sqrt{1 - x^{2}}) \\ On squaring both sides, \\ \Rightarrow 1 - {(1 - x)}^{2} = 4 x^{2} (1 - x^{2}) \\ \Rightarrow 1 - 1 - x^{2} + 2 x = 4 x^{2} - 4 x^{4} \\ \Rightarrow 4 x^{4} - 5 x^{2} + 2 x = 0 \\ \Rightarrow x (4 x^{3} - 5 x + 2) = 0 \\ \Rightarrow x = 0 or \\ 4 x^{3} - 5 x + 2 = 0 \\ Put x = . .. -1, 0, 1, 2, . \dots we get no real vlue of x \\ satisfy above cubic equation in [- 1, 1] \\ Hence x = 0 is only solution . \end{array}$

Q.49

$Solve for x, \tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4} .$

Ans

$\begin{array}{l} We have, \tan^{- 1} 2 x + \tan^{- 1} 2 x = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{2 x + 3 x}{1 - 2 x \times 3 x}) = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{5 x}{1 - 6 x^{2}}) = \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = \tan \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \\ \Rightarrow 5 x = 1 - 6 x^{2} \\ \Rightarrow 6 x^{2} = 5 x - 1 = 0 \\ \Rightarrow (6 x - 1) (x + 1) = 0 \\ \Rightarrow x = - (rejected as does not satisfy equation) \\ \Rightarrow x = \frac{1}{6} is the only solution of the given equation . \end{array}$

Q.50

$Put in the simplest form, \tan^{- 1} (\frac{\cos θ}{1 + \sin θ}) .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{\cos θ}{1 + \sin θ}) \\ = \tan^{- 1} (\frac{\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}}{{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}^{2}}) \\ = \tan^{- 1} (\frac{(\cos \frac{θ}{2} + \sin \frac{θ}{2}) (\cos \frac{θ}{2} - \sin \frac{θ}{2})}{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}) \\ = \tan^{- 1} (\frac{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})}) \\ Divide N^{1} and D^{1} by \cos \frac{θ}{2} \\ = \tan^{- 1} \frac{(1 - \tan \frac{θ}{2})}{(1 + \tan \frac{θ}{2})} \\ = \tan^{- 1} \frac{(\tan \frac{π}{4} - \tan \frac{θ}{2})}{(1 + \tan \frac{π}{4} . \tan \frac{θ}{2})} \\ = \tan^{- 1} [(\tan \frac{π}{4} - \frac{θ}{2})] \\ = \frac{π}{4} - \frac{θ}{2} \end{array}$

Q.51

$Show that \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8} = \frac{π}{4}$

Ans

$\begin{array}{l} We have, \\ \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8} \\ = \tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}) + \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}}) \\ = \tan^{- 1} (\frac{\frac{7 + 5}{35}}{1 - \frac{1}{35}}) + \tan^{- 1} (\frac{\frac{8 + 3}{24}}{1 - \frac{1}{24}}) \\ = \tan^{- 1} (\frac{\frac{12}{35}}{\frac{34}{35}}) + \tan^{- 1} (\frac{\frac{11}{24}}{\frac{23}{24}}) \\ = \tan^{- 1} (\frac{12}{34}) + \tan^{- 1} (\frac{11}{23}) \\ = \tan^{- 1} (\frac{6}{17}) + \tan^{- 1} (\frac{11}{23}) \\ = \tan^{- 1} (\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}}) \\ = \tan^{- 1} (\frac{\frac{138 + 187}{17 \times 23}}{\frac{391 - 66}{17 \times 23}}) \\ = \tan^{- 1} (\frac{325}{325}) \\ = \tan^{- 1} (1) \\ = \frac{π}{4} \\ = 1 \end{array}$

Q.52

$Put in the simplest form, \tan^{- 1} (\frac{\cos x}{1 - \sin x}) .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{\cos x}{1 - \sin x}) \\ = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}) \\ \tan^{- 1} = (\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})}) \\ \tan^{- 1} = (\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})}) \\ Divide N^{1} and D^{1} by \cos \frac{x}{2} \\ = \tan^{- 1} \frac{(1 + \tan \frac{x}{2})}{(1 - \tan \frac{x}{2})} \\ = \tan^{- 1} \frac{(\tan \frac{π}{4} + \tan \frac{x}{2})}{(1 - \tan \frac{π}{4} . \tan \frac{x}{2})} \\ = \tan^{- 1} [(\tan \frac{π}{4} + \frac{x}{2})] \\ = \frac{π}{4} + \frac{x}{2} \end{array}$

Q.53

${Show that tan}^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{3}{4} = \tan^{- 1} 2 .$

Ans

$\begin{array}{l} LHS = \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{1 - \frac{14}{11 \times 24}}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{125}{264 - 14}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{125}{250}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{\frac{1}{2} + \frac{3}{4}}{1 - \frac{1}{2} \times \frac{3}{4}}) \\ = \tan^{- 1} (\frac{\frac{2 + 3}{4}}{\frac{8 - 3}{8}}) \\ = \tan^{- 1} (\frac{\frac{5}{4}}{\frac{5}{8}}) \\ = \tan^{- 1} (\frac{5}{4} \times \frac{8}{5}) \\ = \tan^{- 1} 2 (RHS Proved) \end{array}$

Q.54

$Evaluate \cot (\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8})$

Ans

$\begin{array}{l} We have, \\ \cot [\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}] \\ = \cot [\tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8}] \\ = \cot [\tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}) + \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}})] \\ = \cot [\tan^{- 1} (\frac{\frac{7 + 5}{35}}{1 - \frac{1}{35}}) + \tan^{- 1} (\frac{\frac{8 + 3}{24}}{1 - \frac{1}{24}})] \\ = \cot [\tan^{- 1} (\frac{\frac{12}{35}}{\frac{34}{35}}) + \tan^{- 1} (\frac{\frac{11}{24}}{\frac{23}{24}})] \\ = \cot [\tan^{- 1} (\frac{12}{34}) + \tan^{- 1} (\frac{11}{23})] \\ = \cot [\tan^{- 1} (\frac{6}{17}) + \tan^{- 1} (\frac{11}{23})] \\ = \cot + [\tan^{- 1} (\frac{\frac{138 + 187}{17 \times 23}}{\frac{391 - 66}{17 \times 23}})] \\ = \cot + [\tan^{- 1} (\frac{325}{325})] \\ = \cot + [\tan^{- 1} (1)] \\ = \cot \frac{π}{4} \\ = 1 \end{array}$

Q.55

$\begin{array}{l} If \sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = \frac{π}{2}, then \\ find the value of x^{2} + y^{2} + z^{2} + 2 xyz . \end{array}$

Ans

$\begin{array}{l} We have, \\ \sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = \frac{π}{2} \\ \Rightarrow \cos^{- 1} \sqrt{1 - x^{2}} + \cos^{- 1} \sqrt{1 - y^{2}} = \frac{π}{2} \sin^{- 1} z \\ \Rightarrow \cos^{- 1} [\sqrt{1 - x^{2}} \sqrt{1 - y^{2}} - \sqrt{\{1 - (1 - x^{2})\}} \sqrt{\{1 - (1 - y^{2})\}}] = \cos^{- 1} z \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} - \sqrt{(x^{2})} \sqrt{(y^{2})} = z \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = xy + z \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = xy + z \\ Squaring both sides \\ {(xy + z)}^{2} = (1 - x^{2}) (1 - y^{2}) \\ \Rightarrow x^{2} y^{2} + z^{2} + 2 xyz = 1 - x^{2} - y^{2} + x^{2} y^{2} \\ \Rightarrow x^{2} + y^{2} + z^{2} + 2 xyz = 1 \\ Hence, value of x^{2} + y^{2} + z^{2} + 2 xyz is 1 . \end{array}$

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FAQs (Frequently Asked Questions)

Inverse trigonometric functions need a principal value branch to give one unique answer. Trigonometric functions repeat their values, so the same input can have many possible angles. A fixed branch removes this ambiguity.

Group them in pairs. sin⁻¹x and tan⁻¹x have ranges around 0. cos⁻¹x and cot⁻¹x have positive ranges. sec⁻¹x follows the cosine range with π/2 removed, while cosec⁻¹x follows the sine range with 0 removed.

cot⁻¹x has the principal range (0, π). Although cot(-π/4) = -1, -π/4 does not lie in this range. The angle 3π/4 lies in (0, π), so cot⁻¹(-1) = 3π/4.

sin⁻¹(sin x) is equal to x only when x lies in [-π/2, π/2]. If x lies outside this interval, it must be converted to an equivalent angle within the principal value branch.

Conditions decide the correct principal value. For tan⁻¹x + tan⁻¹y, the formula changes depending on xy. If the condition is ignored, the answer may fall outside the correct principal branch.

CBSE Class 12 Maths Revision Notes Chapter 2 Inverse Trigonometric Functions

Key Takeaways

Access Class 12 Maths Chapter 2 Inverse Trigonometric Functions Notes in 30 Minutes

Inverse Trigonometric Functions Class 12 Notes: Basic Concept

Why Trigonometric Functions Need Restricted Domains

Principal Value Branch

Domain and Range of Inverse Trigonometric Functions

sin⁻¹x: Domain, Range and Principal Value Branch

cos⁻¹x: Domain, Range and Principal Value Branch

tan⁻¹x: Domain, Range and Principal Value Branch

cot⁻¹x: Domain, Range and Principal Value Branch

sec⁻¹x: Domain, Range and Principal Value Branch

cosec⁻¹x: Domain, Range and Principal Value Branch

Graphs of Inverse Trigonometric Functions

Properties of Inverse Trigonometric Functions

Reciprocal Identities of Inverse Trigonometric Functions

Negative Argument Identities

Complementary Identities

Addition and Subtraction Formulas

Double Angle Formulas in Inverse Trigonometry

Principal Value Examples for Quick Revision

Example 1: Find sin⁻¹(1/2)

Example 2: Find cos⁻¹(-1/2)

Example 3: Find tan⁻¹(-1)

Example 4: Find cot⁻¹(-1)

Example 5: Find sec⁻¹(2)

Example 6: Find cosec⁻¹(-2)

Simplifying Inverse Trigonometric Expressions

Standard Simplification Results

Remember These Points Before Solving

Common Mistakes in Class 12 Mathematics Chapter 2 Notes

Quick Revision Summary for Inverse Trigonometric Functions

Important Terms in Inverse Trigonometric Functions

Useful Links for Class 12 Maths

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FAQs (Frequently Asked Questions)

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CBSE Class 12 Maths Revision Notes Chapter 2 Inverse Trigonometric Functions

Key Takeaways

Access Class 12 Maths Chapter 2 Inverse Trigonometric Functions Notes in 30 Minutes

Inverse Trigonometric Functions Class 12 Notes: Basic Concept

Why Trigonometric Functions Need Restricted Domains

Principal Value Branch

Domain and Range of Inverse Trigonometric Functions

sin⁻¹x: Domain, Range and Principal Value Branch

cos⁻¹x: Domain, Range and Principal Value Branch

tan⁻¹x: Domain, Range and Principal Value Branch

cot⁻¹x: Domain, Range and Principal Value Branch

sec⁻¹x: Domain, Range and Principal Value Branch

cosec⁻¹x: Domain, Range and Principal Value Branch

Graphs of Inverse Trigonometric Functions

Properties of Inverse Trigonometric Functions

Reciprocal Identities of Inverse Trigonometric Functions

Negative Argument Identities

Complementary Identities

Addition and Subtraction Formulas

Double Angle Formulas in Inverse Trigonometry

Principal Value Examples for Quick Revision

Example 1: Find sin⁻¹(1/2)

Example 2: Find cos⁻¹(-1/2)

Example 3: Find tan⁻¹(-1)

Example 4: Find cot⁻¹(-1)

Example 5: Find sec⁻¹(2)

Example 6: Find cosec⁻¹(-2)

Simplifying Inverse Trigonometric Expressions

Standard Simplification Results

Remember These Points Before Solving

Common Mistakes in Class 12 Mathematics Chapter 2 Notes

Quick Revision Summary for Inverse Trigonometric Functions

Important Terms in Inverse Trigonometric Functions

Useful Links for Class 12 Maths

Please register to view this section

FAQs (Frequently Asked Questions)

1. Why do inverse trigonometric functions need a principal value branch?

2. How can I remember the range of all inverse trigonometric functions?

3. Why is cot⁻¹(-1) equal to 3π/4 and not -π/4?

4. When is sin⁻¹(sin x) equal to x?

5. Why should conditions be checked in tan⁻¹ addition formulas?

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