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Inverse Trigonometric Functions

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Learning Objectives

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Learning Objectives

  • Understand the concept of inverse functions and their existence conditions.
  • Identify the restrictions on domains and ranges of trigonometric functions for defining their inverses.
  • Analyze the graphical representations of inverse trigonometric functions.
  • Recognize the principal value branches of inverse trigonometric functions.
  • Apply properties of inverse trigonometric functions in solving equations.
  • Differentiate between inverse trigonometric functions and their reciprocal functions.

Revision Notes & Summary

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Inverse Trigonometric Functions

2.1 Introduction

  • The inverse of a function f, denoted by f⁻¹, exists if f is one-one and onto.
  • Trigonometric functions are not one-one and onto over their natural domains and ranges; hence their inverses do not exist without restrictions.
  • This chapter discusses restrictions on domains and ranges of trigonometric functions to ensure the existence of their inverses.
  • The inverse trigonometric functions are important in calculus and have applications in science and engineering.

2.2 Basic Concepts

Trigonometric Functions

  • Sine Function:
    • Domain: R
    • Range: [-1, 1]
  • Cosine Function:
    • Domain: R
    • Range: [-1, 1]
  • Tangent Function:
    • Domain: R ackslash {x: x = (2n + 1)π, n ∈ Z}
    • Range: R
  • Cotangent Function:
    • Domain: R ackslash {x: x = nπ, n ∈ Z}
    • Range: R
  • Secant Function:
    • Domain: R ackslash {x: x = (2n + π/2), n ∈ Z}
    • Range: R ackslash (-1, 1)
  • Cosecant Function:
    • Domain: R ackslash {x: x = nπ, n ∈ Z}
    • Range: R ackslash (-1, 1)

2.3 Properties of Inverse Trigonometric Functions

  • The principal value branch of inverse trigonometric functions is defined as follows:
    • If y = sin⁻¹ x, then x = sin y and vice versa.
    • The principal value of sin⁻¹ x lies in the range of [-π/2, π/2].
    • Similar definitions apply for other inverse trigonometric functions.

Example Problems

  1. Find the principal value of sin⁻¹(1/2):
    • Solution: sin⁻¹(1/2) = π/6
  2. Find the principal value of cot⁻¹(1):
    • Solution: cot⁻¹(1) = π/4

2.4 Summary of Domains and Ranges

FunctionsDomainRange (Principal Value Branches)
y = sin⁻¹ x[-1, 1][-π/2, π/2]
y = cos⁻¹ x[-1, 1][0, π]
y = cosec⁻¹ xR ackslash (-1, 1)[-π/2, 0) ∪ (0, π/2]
y = sec⁻¹ xR ackslash (-1, 1)[0, π/2) ∪ (π/2, π]
y = tan⁻¹ xR(-π/2, π/2)
y = cot⁻¹ xR(0, π)

Exam Tips & Common Mistakes

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Common Mistakes and Exam Tips

Common Pitfalls

  • Confusing Inverse Functions: Students often confuse inverse trigonometric functions (e.g., sin⁻¹x) with the reciprocal of trigonometric functions (e.g., (sin x)⁻¹). Remember that sin⁻¹x is the angle whose sine is x, while (sin x)⁻¹ is the cosecant of x.
  • Ignoring Domain Restrictions: Inverse trigonometric functions have specific domains and ranges. For example, sin⁻¹x is defined for x in [-1, 1] and its range is [-π/2, π/2]. Not adhering to these can lead to incorrect answers.
  • Misapplying Principal Values: When asked for the principal value of an inverse function, ensure you are within the specified range. For instance, the principal value of sin⁻¹(1/2) is π/6, not any other angle that has the same sine value.
  • Incorrectly Solving Equations: When solving equations involving inverse trigonometric functions, ensure to apply the correct identities and properties. For example, if sin⁻¹x = y, then sin(y) = x must hold true.

Tips for Exam Success

  • Review Graphs: Familiarize yourself with the graphs of inverse trigonometric functions, as they can help visualize the domain and range.
  • Practice with Examples: Work through various examples to solidify your understanding of how to find principal values and solve equations involving inverse trigonometric functions.
  • Check Your Work: After solving problems, check if your answers fall within the expected ranges for inverse functions. This can help catch mistakes early.
  • Understand Properties: Make sure to understand the properties of inverse trigonometric functions, such as sin(sin⁻¹x) = x for x in [-1, 1]. This can be useful for verification.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

00

B.

π\pi

C.

−π-\pi

D.

π2\frac{\pi}{2}
Correct Answer: B

Solution:

The principal value of cos⁡−1(−1)\cos^{-1}(-1) is π\pi, as the range of cos⁡−1(x)\cos^{-1}(x) is [0,π][0, \pi] and cos⁡(π)=−1\cos(\pi) = -1.

A.

cos⁡−1(1x)\cos^{-1}(\frac{1}{x})

B.

sin⁡−1(1x)\sin^{-1}(\frac{1}{x})

C.

tan⁡−1(1x)\tan^{-1}(\frac{1}{x})

D.

csc⁡−1(1x)\csc^{-1}(\frac{1}{x})
Correct Answer: A

Solution:

For x>1x > 1, sec⁡−1(x)=cos⁡−1(1x)\sec^{-1}(x) = \cos^{-1}(\frac{1}{x}) because the secant function is the reciprocal of the cosine function. The range of sec⁡−1(x)\sec^{-1}(x) is [0,π][0, \pi] excluding π2\frac{\pi}{2}, which matches the range of cos⁡−1(1x)\cos^{-1}(\frac{1}{x}) for x>1x > 1.

A.

(−∞,∞)(-\infty, \infty)

B.

[0,π][0, \pi]

C.

[−1,1][-1, 1]

D.

(0,π)(0, \pi)
Correct Answer: A

Solution:

The domain of the cosine function, cos⁡(x)\cos(x), is all real numbers, (−∞,∞)(-\infty, \infty).

A.

π3\frac{\pi}{3}

B.

π4\frac{\pi}{4}

C.

π6\frac{\pi}{6}

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of tan⁡−1(x)\tan^{-1}(x) is in the range (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). For x=3x = \sqrt{3}, tan⁡−1(3)=π3\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.

A.

−π3-\frac{\pi}{3}

B.

π3\frac{\pi}{3}

C.

−π6-\frac{\pi}{6}

D.

π6\frac{\pi}{6}
Correct Answer: A

Solution:

The principal value of tan⁡−1(−3)\tan^{-1}(-\sqrt{3}) is −π3-\frac{\pi}{3}, as the range of the principal value branch of tan⁡−1\tan^{-1} is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

A.

x=10x = \frac{1}{0}

B.

x=−3x = -\sqrt{3}

C.

x=π2x = \frac{\pi}{2}

D.

x=2πx = 2\pi
Correct Answer: B

Solution:

The function tan⁡−1(x)\tan^{-1}(x) is defined for all real numbers xx. Therefore, x=−3x = -\sqrt{3} is a valid input for tan⁡−1(x)\tan^{-1}(x).

A.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

B.

(0,π)(0, \pi)

C.

[0,π][0, \pi]

D.

(−π,π)(-\pi, \pi)
Correct Answer: A

Solution:

The principal value branch of the inverse sine function, sin⁡−1x\sin^{-1} x, has a range of (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

A.

11−x2\frac{1}{\sqrt{1-x^2}}

B.

−11−x2\frac{-1}{\sqrt{1-x^2}}

C.

1−x2\sqrt{1-x^2}

D.

−1−x2-\sqrt{1-x^2}
Correct Answer: A

Solution:

The derivative of y=sin⁡−1(x)y = \sin^{-1}(x) with respect to xx is 11−x2\frac{1}{\sqrt{1-x^2}}, valid for x∈(−1,1)x \in (-1, 1).

A.

π4\frac{\pi}{4}

B.

π2\frac{\pi}{2}

C.

π\pi

D.

00
Correct Answer: A

Solution:

The principal value of cot⁡−1(1)\cot^{-1}(1) is π4\frac{\pi}{4} because cot⁡(π4)=1\cot(\frac{\pi}{4}) = 1 and it lies in the range (0,π)(0, \pi) for cot⁡−1(x)\cot^{-1}(x).

A.

π3\frac{\pi}{3}

B.

π2\frac{\pi}{2}

C.

π\pi

D.

00
Correct Answer: D

Solution:

The range of the principal value branch of cot⁡−1(x)\cot^{-1}(x) is (0,π)(0, \pi), so 00 is not a valid principal value.

A.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

B.

(0,π)(0, \pi)

C.

(−π,π)(-\pi, \pi)

D.

(0,2π)(0, 2\pi)
Correct Answer: A

Solution:

The range of the principal value branch of tan⁡−1(x)\tan^{-1}(x) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), which is where the function is defined.

A.

π2\frac{\pi}{2}

B.

π\pi

C.

00

D.

3π2\frac{3\pi}{2}
Correct Answer: A

Solution:

tan⁡(x)\tan(x) is undefined at x=π2+nπx = \frac{\pi}{2} + n\pi, where nn is an integer. Thus, x=π2x = \frac{\pi}{2} is a possible value.

A.

(−∞,∞)(-\infty, \infty)

B.

[0,π][0, \pi]

C.

[−1,1][-1, 1]

D.

[0,2π][0, 2\pi]
Correct Answer: A

Solution:

The cosine function, cos⁡(x)\cos(x), is defined for all real numbers, hence its domain is (−∞,∞)(-\infty, \infty).

A.

Its range is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

B.

Its range is [0,π][0, \pi].

C.

Its domain is (−1,1)(-1, 1).

D.

Its domain is [0,∞)[0, \infty).
Correct Answer: A

Solution:

The range of the inverse tangent function, tan⁡−1x\tan^{-1} x, is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

A.

π6\frac{\pi}{6}

B.

π3\frac{\pi}{3}

C.

π2\frac{\pi}{2}

D.

π4\frac{\pi}{4}
Correct Answer: A

Solution:

The principal value of cot⁡−1(3)\cot^{-1}(\sqrt{3}) is π6\frac{\pi}{6} because cot⁡(π6)=3\cot(\frac{\pi}{6}) = \sqrt{3}. The range of the principal value of cot⁡−1\cot^{-1} is (0,π)(0, \pi).

A.

π2\frac{\pi}{2}

B.

π\pi

C.

3π2\frac{3\pi}{2}

D.

2π2\pi
Correct Answer: B

Solution:

tan⁡(x)\tan(x) is undefined at odd multiples of π2\frac{\pi}{2}, so x=πx = \pi is a possible value where tan⁡(x)\tan(x) is defined.

A.

00

B.

π\pi

C.

−π-\pi

D.

π2\frac{\pi}{2}
Correct Answer: B

Solution:

The principal value of cos⁡−1(−1)\cos^{-1}(-1) is π\pi because the range of the principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi].

A.

π6\frac{\pi}{6}

B.

π3\frac{\pi}{3}

C.

π4\frac{\pi}{4}

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of sin⁡−1(12)\sin^{-1}(\frac{1}{2}) is π6\frac{\pi}{6} because sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2} and it lies within the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

[0,π][0, \pi] excluding π2\frac{\pi}{2}

B.

[0,π][0, \pi]

C.

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

D.

[0,π2][0, \frac{\pi}{2}]
Correct Answer: A

Solution:

The range of the principal value of y=sec⁡−1(x)y = \sec^{-1}(x) is [0,π][0, \pi] excluding π2\frac{\pi}{2}.

A.

The inverse of a trigonometric function exists only if it is one-one.

B.

The inverse of a trigonometric function exists only if it is onto.

C.

The inverse of a trigonometric function exists only if it is both one-one and onto.

D.

The inverse of a trigonometric function exists regardless of being one-one or onto.
Correct Answer: C

Solution:

The inverse of a function exists if and only if the function is both one-one and onto.

A.

π6\frac{\pi}{6}

B.

π4\frac{\pi}{4}

C.

π3\frac{\pi}{3}

D.

π2\frac{\pi}{2}
Correct Answer: C

Solution:

The principal value of csc⁡−1(x)\csc^{-1}(x) is in the range [−π2,π2]∖{0}[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}. For x=2x = 2, csc⁡−1(2)=π3\csc^{-1}(2) = \frac{\pi}{3}.

A.

11+x2\frac{1}{1 + x^2}

B.

11−x2\frac{1}{1 - x^2}

C.

1x2−1\frac{1}{x^2 - 1}

D.

x1+x2\frac{x}{1 + x^2}
Correct Answer: A

Solution:

The derivative of f(x)=tan⁡−1(x)f(x) = \tan^{-1}(x) is f′(x)=11+x2f'(x) = \frac{1}{1 + x^2}. This is derived from the definition of the derivative of the inverse tangent function.

A.

x=yx = y

B.

x=−yx = -y

C.

x=1−y1+yx = \frac{1-y}{1+y}

D.

x=1+y1−yx = \frac{1+y}{1-y}
Correct Answer: C

Solution:

Using the identity tan⁡−1(x)+tan⁡−1(y)=tan⁡−1(x+y1−xy)\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right), and given tan⁡−1(x)+tan⁡−1(y)=π4\tan^{-1}(x) + \tan^{-1}(y) = \frac{\pi}{4}, we have x+y1−xy=1\frac{x+y}{1-xy} = 1. Solving this equation gives x+y=1−xyx+y = 1-xy, which simplifies to x=1−y1+yx = \frac{1-y}{1+y}.

A.

R∖{x:x=(2n+1)π2,n∈Z}R \setminus \{ x : x = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} \}

B.

R∖{x:x=nπ,n∈Z}R \setminus \{ x : x = n\pi, n \in \mathbb{Z} \}

C.

R∖{x:x=(2n+1)π,n∈Z}R \setminus \{ x : x = (2n+1)\pi, n \in \mathbb{Z} \}

D.

R∖{x:x=nπ2,n∈Z}R \setminus \{ x : x = n\frac{\pi}{2}, n \in \mathbb{Z} \}
Correct Answer: A

Solution:

The domain of sec⁡(x)\sec(x) is R∖{x:x=(2n+1)π2,n∈Z}R \setminus \{ x : x = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} \} because sec⁡(x)\sec(x) is undefined where cos⁡(x)=0\cos(x) = 0.

A.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

B.

[0,π][0, \pi]

C.

(−π,π)(-\pi, \pi)

D.

[0,π2][0, \frac{\pi}{2}]
Correct Answer: A

Solution:

The principal value branch of sin⁡−1(x)\sin^{-1}(x) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) because this is the range where the inverse sine function is defined to be one-one and onto.

A.

1−x2\sqrt{1-x^2}

B.

1−x21-x^2

C.

x2x^2

D.

xx
Correct Answer: A

Solution:

For xx in the interval [0,1][0, 1], cos⁡−1(x)\cos^{-1}(x) is the angle whose cosine is xx. Therefore, sin⁡(cos⁡−1(x))=1−x2\sin(\cos^{-1}(x)) = \sqrt{1-x^2}, because sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.

A.

(−∞,∞)(-\infty, \infty)

B.

[−1,1][-1, 1]

C.

(0,π)(0, \pi)

D.

[0,π][0, \pi]
Correct Answer: B

Solution:

The domain of the inverse cosine function, cos⁡−1(x)\cos^{-1}(x), is [−1,1][-1, 1].

A.

−π2-\frac{\pi}{2}

B.

3π2\frac{3\pi}{2}

C.

π\pi

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of sin⁡−1(−1)\sin^{-1}(-1) is −π2-\frac{\pi}{2} because the range of the principal value branch of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

x=(2n+1)π2,n∈Zx = (2n + 1)\frac{\pi}{2}, n \in \mathbb{Z}

B.

x=nπ,n∈Zx = n\pi, n \in \mathbb{Z}

C.

x=(2n)π2,n∈Zx = (2n)\frac{\pi}{2}, n \in \mathbb{Z}

D.

x=(2n+1)π,n∈Zx = (2n + 1)\pi, n \in \mathbb{Z}
Correct Answer: A

Solution:

The tangent function, tan⁡(x)\tan(x), is undefined at x=(2n+1)π2,n∈Zx = (2n + 1)\frac{\pi}{2}, n \in \mathbb{Z}, because these are the points where the cosine function, which is in the denominator, equals zero.

A.

The domain of sec⁡−1(x)\sec^{-1}(x) is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty).

B.

The range of sec⁡−1(x)\sec^{-1}(x) is [0,π][0, \pi] excluding π2\frac{\pi}{2}.

C.

The function sec⁡−1(x)\sec^{-1}(x) is undefined for x=0x = 0.

D.

The domain of sec⁡−1(x)\sec^{-1}(x) is (−1,1)(-1, 1).
Correct Answer: A

Solution:

The domain of sec⁡−1(x)\sec^{-1}(x) is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty) because the secant function is undefined between −1-1 and 11.

A.

[0,π][0, \pi]

B.

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

C.

(−∞,∞)(-\infty, \infty)

D.

[0,π2][0, \frac{\pi}{2}]
Correct Answer: B

Solution:

The range of the principal value branch of sin⁡−1(x)\sin^{-1}(x) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

−11+x2-\frac{1}{1 + x^2}

B.

11+x2\frac{1}{1 + x^2}

C.

−11−x2-\frac{1}{1 - x^2}

D.

11−x2\frac{1}{1 - x^2}
Correct Answer: A

Solution:

The derivative of y=cot⁡−1(x)y = \cot^{-1}(x) with respect to xx is −11+x2-\frac{1}{1 + x^2}.

A.

x=2x = 2

B.

x=12x = \frac{1}{2}

C.

x=−2x = -2

D.

x=3x = \sqrt{3}
Correct Answer: A

Solution:

The principal value of sec⁡−1(x)\sec^{-1}(x) is π3\frac{\pi}{3} when x=2x = 2. This is because sec⁡(π3)=2\sec(\frac{\pi}{3}) = 2.

A.

3π4\frac{3\pi}{4}

B.

π4\frac{\pi}{4}

C.

π2\frac{\pi}{2}

D.

π\pi
Correct Answer: A

Solution:

The principal value of cot⁡−1(x)\cot^{-1}(x) is in the range (0,π)(0, \pi). For x=−1x = -1, cot⁡−1(−1)=3π4\cot^{-1}(-1) = \frac{3\pi}{4}.

A.

−π2-\frac{\pi}{2}

B.

π\pi

C.

−π-\pi

D.

00
Correct Answer: A

Solution:

The principal value of sin⁡−1(−1)\sin^{-1}(-1) is −π2-\frac{\pi}{2} because the range of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

(0,π)(0, \pi)

B.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

C.

[0,π][0, \pi]

D.

[0,π2)[0, \frac{\pi}{2})
Correct Answer: A

Solution:

The range of the principal value branch of cot⁡−1(x)\cot^{-1}(x) is (0,π)(0, \pi).

A.

3π2\frac{3\pi}{2}

B.

π\pi

C.

π4\frac{\pi}{4}

D.

−π-\pi
Correct Answer: C

Solution:

The range of the principal value branch of tan⁡−1(x)\tan^{-1}(x) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), so π4\frac{\pi}{4} is within this range.

A.

−π2-\frac{\pi}{2}

B.

π\pi

C.

00

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of sin⁡−1(−1)\sin^{-1}(-1) is −π2-\frac{\pi}{2} because the range of the principal value branch of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

B.

(0,π)(0, \pi)

C.

(−π,π)(-\pi, \pi)

D.

(0,2π)(0, 2\pi)
Correct Answer: A

Solution:

The principal value branch of sin⁡−1(x)\sin^{-1}(x) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), which is the range where the function is defined.

A.

π6\frac{\pi}{6}

B.

π3\frac{\pi}{3}

C.

π2\frac{\pi}{2}

D.

π\pi
Correct Answer: A

Solution:

The principal value of sin⁡−1(12)\sin^{-1}(\frac{1}{2}) is π6\frac{\pi}{6} because sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}.

A.

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

B.

[0,π][0, \pi]

C.

(−∞,∞)(-\infty, \infty)

D.

[0,π2][0, \frac{\pi}{2}]
Correct Answer: A

Solution:

The principal value branch of the inverse sine function sin⁡−1(x)\sin^{-1}(x) is defined in the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], which ensures the function is one-one and onto.

A.

−π6-\frac{\pi}{6}

B.

π6\frac{\pi}{6}

C.

−π4-\frac{\pi}{4}

D.

π4\frac{\pi}{4}
Correct Answer: A

Solution:

The principal value of sin⁡−1(x)\sin^{-1}(x) is in the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. For x=−12x = -\frac{1}{2}, sin⁡−1(−12)=−π6\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}.

A.

π3\frac{\pi}{3}

B.

π\pi

C.

π2\frac{\pi}{2}

D.

2π3\frac{2\pi}{3}
Correct Answer: C

Solution:

The principal value branch for sec⁡−1(x)\sec^{-1}(x) is [0,π][0, \pi] excluding π2\frac{\pi}{2}, so π2\frac{\pi}{2} is not a valid principal value.

A.

−12-\frac{1}{2}

B.

12\frac{1}{2}

C.

−32-\frac{\sqrt{3}}{2}

D.

32\frac{\sqrt{3}}{2}
Correct Answer: A

Solution:

The cosine of 2π3\frac{2\pi}{3} is −12-\frac{1}{2}, thus x=−12x = -\frac{1}{2}.

A.

−π3-\frac{\pi}{3}

B.

−π4-\frac{\pi}{4}

C.

−π6-\frac{\pi}{6}

D.

−π2-\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of tan⁡−1(−3)\tan^{-1}(-\sqrt{3}) is −π3-\frac{\pi}{3} because the range of the principal value branch of tan⁡−1\tan^{-1} is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

A.

xy=1xy = 1

B.

x+y=1x + y = 1

C.

x−y=1x - y = 1

D.

xy=−1xy = -1
Correct Answer: A

Solution:

Using the identity tan⁡−1(x)+tan⁡−1(y)=tan⁡−1(x+y1−xy)\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right), and given tan⁡−1(x)+tan⁡−1(y)=π4\tan^{-1}(x) + \tan^{-1}(y) = \frac{\pi}{4}, we have tan⁡(π4)=1\tan\left(\frac{\pi}{4}\right) = 1, thus xy=1xy = 1.

A.

x=nπ,n∈Zx = n\pi, n \in \mathbb{Z}

B.

x=(2n+1)π2,n∈Zx = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}

C.

x=(2n+1)π,n∈Zx = (2n+1)\pi, n \in \mathbb{Z}

D.

x=2nπ,n∈Zx = 2n\pi, n \in \mathbb{Z}
Correct Answer: B

Solution:

The secant function sec⁡(x)\sec(x) is undefined for x=(2n+1)π2,n∈Zx = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} because cos⁡(x)=0\cos(x) = 0 at these points.

A.

[0,π][0, \pi]

B.

[−π,π][-\pi, \pi]

C.

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

D.

[0,2π][0, 2\pi]
Correct Answer: A

Solution:

The range of cos⁡−1x\cos^{-1} x is [0,π][0, \pi].

A.

22

B.

12\frac{1}{2}

C.

13\frac{1}{\sqrt{3}}

D.

232\sqrt{3}
Correct Answer: A

Solution:

The principal value of sec⁡−1(x)\sec^{-1}(x) is π3\frac{\pi}{3} when x=2x = 2, because sec⁡(π3)=2\sec(\frac{\pi}{3}) = 2.

A.

−π2-\frac{\pi}{2}

B.

π\pi

C.

−π-\pi

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of sin⁡−1(−1)\sin^{-1}(-1) is −π2-\frac{\pi}{2} because the range of the principal value branch of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

π6\frac{\pi}{6}

B.

π4\frac{\pi}{4}

C.

π3\frac{\pi}{3}

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of csc⁡−1(x)\csc^{-1}(x) is the angle θ\theta such that csc⁡(θ)=x\csc(\theta) = x and θ\theta is in the range [−π2,π2]∖{0}[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}. For csc⁡−1(2)\csc^{-1}(2), θ=π6\theta = \frac{\pi}{6} since csc⁡(π6)=2\csc(\frac{\pi}{6}) = 2.

A.

x=2x = 2

B.

x=12x = \frac{1}{2}

C.

x=3x = \sqrt{3}

D.

x=13x = \frac{1}{\sqrt{3}}
Correct Answer: A

Solution:

The principal value of csc⁡−1(x)\csc^{-1}(x) is π6\frac{\pi}{6} when x=2x = 2, since csc⁡(π6)=2\csc(\frac{\pi}{6}) = 2.

A.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

B.

[0,π][0, \pi]

C.

(−π,π)(-\pi, \pi)

D.

[0,π2][0, \frac{\pi}{2}]
Correct Answer: A

Solution:

The principal value branch of tan⁡−1(x)\tan^{-1}(x) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), which ensures the function is one-one and onto in this interval.

A.

It is defined for all real numbers.

B.

It is only defined for x≤−1x \leq -1 or x≥1x \geq 1.

C.

It is only defined for −1<x<1-1 < x < 1.

D.

It is defined for x=0x = 0.
Correct Answer: B

Solution:

The function sec⁡−1\sec^{-1} is defined for x≤−1x \leq -1 or x≥1x \geq 1 because sec⁡x\sec x cannot take values between −1-1 and 11.

A.

3π4\frac{3\pi}{4}

B.

π\pi

C.

π2\frac{\pi}{2}

D.

00
Correct Answer: D

Solution:

The principal value branch of cot⁡−1(x)\cot^{-1}(x) is (0,π)(0, \pi), so 00 is within this range.

A.

(−∞,∞)(-\infty, \infty)

B.

[−1,1][-1, 1]

C.

(0,π)(0, \pi)

D.

[0,π][0, \pi]
Correct Answer: B

Solution:

The domain of the inverse cosine function, cos⁡−1x\cos^{-1} x, is [−1,1][-1, 1].

A.

x∈Rx \in \mathbb{R}

B.

x∈[0,π]x \in [0, \pi]

C.

x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

D.

x∈[0,2π]x \in [0, 2\pi]
Correct Answer: A

Solution:

The domain of the sine function sin⁡(x)\sin(x) is all real numbers x∈Rx \in \mathbb{R}.

A.

12\frac{1}{2}

B.

32\frac{\sqrt{3}}{2}

C.

−12-\frac{1}{2}

D.

−32-\frac{\sqrt{3}}{2}
Correct Answer: B

Solution:

The value of xx is 12\frac{1}{2} because cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}.

A.

π2\frac{\pi}{2}

B.

π\pi

C.

00

D.

3π2\frac{3\pi}{2}
Correct Answer: A

Solution:

The identity sin⁡−1(x)+cos⁡−1(x)=π2\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} holds for all xx in the domain [−1,1][-1, 1]. This is because the sine and cosine functions are complementary angles.

A.

(0,π)(0, \pi)

B.

(−π,0)(-\pi, 0)

C.

(0,π2)(0, \frac{\pi}{2})

D.

(π2,π)(\frac{\pi}{2}, \pi)
Correct Answer: A

Solution:

The range of the principal value branch of cot⁡−1\cot^{-1} is (0,π)(0, \pi).

A.

π6\frac{\pi}{6}

B.

π3\frac{\pi}{3}

C.

π4\frac{\pi}{4}

D.

π2\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of csc⁡−1(2)\csc^{-1}(2) is π6\frac{\pi}{6}, as csc⁡(π6)=2\csc(\frac{\pi}{6}) = 2 and the range of csc⁡−1\csc^{-1} is [−π2,π2]∖{0}[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}.

A.

xy=2xy = 2

B.

xy=1xy = 1

C.

x+y=1x + y = 1

D.

x−y=1x - y = 1
Correct Answer: B

Solution:

The identity tan⁡−1(x)+tan⁡−1(y)=π4\tan^{-1}(x) + \tan^{-1}(y) = \frac{\pi}{4} implies xy=1xy = 1 when both xx and yy are positive. This is derived from the tangent addition formula.

A.

11

B.

−1-1

C.

00

D.

3\sqrt{3}
Correct Answer: A

Solution:

The value of xx is 11 because tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1.

A.

Range: (−π,π)(-\pi, \pi)

B.

Range: [0,π][0, \pi]

C.

Range: (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

D.

Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
Correct Answer: D

Solution:

The range of the inverse tangent function tan⁡−1(x)\tan^{-1}(x) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

The range is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

B.

The domain is (−∞,∞)(-\infty, \infty).

C.

The range is [0,π][0, \pi].

D.

The domain is [0,1][0, 1].
Correct Answer: A

Solution:

The range of the inverse sine function, sin⁡−1(x)\sin^{-1}(x), is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

00

B.

π\pi

C.

π2\frac{\pi}{2}

D.

−π-\pi
Correct Answer: B

Solution:

The principal value of cos⁡−1(−1)\cos^{-1}(-1) is π\pi because cos⁡(π)=−1\cos(\pi) = -1.

A.

π2\frac{\pi}{2}

B.

π\pi

C.

00

D.

2π2\pi
Correct Answer: A

Solution:

The sum of sin⁡−1(x)\sin^{-1}(x) and cos⁡−1(x)\cos^{-1}(x) is always π2\frac{\pi}{2} for any xx in the domain [−1,1][-1, 1].

A.

Domain: [−1,1][-1, 1], Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

B.

Domain: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], Range: [−1,1][-1, 1]

C.

Domain: [0,π][0, \pi], Range: [−1,1][-1, 1]

D.

Domain: [−1,1][-1, 1], Range: [0,π][0, \pi]
Correct Answer: A

Solution:

The inverse sine function sin⁡−1(x)\sin^{-1}(x) is defined for xx in the interval [−1,1][-1, 1] and its range is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

A.

(0,π)(0, \pi)

B.

(0,π2)(0, \frac{\pi}{2})

C.

(π2,π)(\frac{\pi}{2}, \pi)

D.

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})
Correct Answer: A

Solution:

The range of the principal value branch of cot⁡−1\cot^{-1} is (0,π)(0, \pi).

A.

−π3-\frac{\pi}{3}

B.

−π4-\frac{\pi}{4}

C.

−π6-\frac{\pi}{6}

D.

−π2-\frac{\pi}{2}
Correct Answer: A

Solution:

The principal value of tan⁡−1(−3)\tan^{-1}(-\sqrt{3}) is −π3-\frac{\pi}{3} because the range of tan⁡−1\tan^{-1} is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

A.

The range of sin⁡−1(x)\sin^{-1}(x) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

B.

The domain of sin⁡−1(x)\sin^{-1}(x) is (−∞,∞)(-\infty, \infty).

C.

The function sin⁡−1(x)\sin^{-1}(x) is defined for all real numbers.

D.

The range of sin⁡−1(x)\sin^{-1}(x) is [0,π][0, \pi].
Correct Answer: A

Solution:

The range of the inverse sine function sin⁡−1(x)\sin^{-1}(x) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

True or False

Correct Answer: True

Solution:

The sine function maps real numbers to the interval [−1,1][-1, 1].

Correct Answer: True

Solution:

The principal value branch of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], and sin⁡−1(−1)=−π2\sin^{-1}(-1) = -\frac{\pi}{2}.

Correct Answer: False

Solution:

Whenever no branch of an inverse trigonometric function is mentioned, it is assumed to be the principal value branch.

Correct Answer: False

Solution:

The inverse sine function, sin⁡−1x\sin^{-1}x, is only defined for xx in the interval [−1,1][-1, 1].

Correct Answer: True

Solution:

The notation y=cos⁡−1xy = \cos^{-1} x is used to denote the inverse cosine function.

Correct Answer: True

Solution:

Inverse trigonometric functions play an important role in calculus as they are used to define many integrals.

Correct Answer: True

Solution:

The range of the principal value branch of sin⁡−1x\sin^{-1} x is indeed [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Correct Answer: False

Solution:

The sine function is not one-one and onto over its natural domain mathbbR\\mathbb{R}; it needs domain restriction to have an inverse.

Correct Answer: True

Solution:

The principal value branch for the inverse sine function is indeed the interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], where it is one-one and onto.

Correct Answer: False

Solution:

The function cos⁡−1x\cos^{-1} x does not have a sinusoidal graph; it is a decreasing function on its domain.

Correct Answer: True

Solution:

A function must be both one-one and onto for its inverse to exist. This is a fundamental property of inverse functions.

Correct Answer: False

Solution:

The notation sin⁻¹x refers to the inverse sine function, while (sin x)⁻¹ refers to the reciprocal of the sine function.

Correct Answer: True

Solution:

A function must be both one-one and onto for its inverse to exist.

Correct Answer: False

Solution:

The sine function is not one-one and onto over its natural domain, which is why its inverse does not exist without restrictions.

Correct Answer: True

Solution:

The principal value is defined as the value of an inverse trigonometric function that lies within the range of its principal branch.

Correct Answer: False

Solution:

In Class XI, it was studied that trigonometric functions, including the sine function, are not one-one and onto over their natural domains and ranges.

Correct Answer: True

Solution:

The tangent function has vertical asymptotes at x=(2n+1)πx = (2n + 1)\pi because it is undefined at these points.

Correct Answer: True

Solution:

The range of the principal value branch of cot⁡−1x\cot^{-1} x is indeed (0,π)(0, \pi).

Correct Answer: False

Solution:

The cosine function is not one-one and onto over its natural domain and range, which is why restrictions are applied to define its inverse.

Correct Answer: True

Solution:

A function must be both one-one and onto for its inverse to exist. This is a fundamental property of functions.

Correct Answer: False

Solution:

The notation sin⁻¹x refers to the inverse sine function, while (sin x)⁻¹ refers to the reciprocal of the sine function.

Correct Answer: False

Solution:

The graph of the inverse cosine function does not oscillate; it is a monotonic decreasing function from 1 to -1.

Correct Answer: False

Solution:

Inverse trigonometric functions require restrictions on their domains and ranges to ensure they are one-one and onto, allowing their inverses to exist.

Correct Answer: True

Solution:

The principal value branch of the cotangent inverse function is defined over the range (0, π).

Correct Answer: False

Solution:

The graph of y=cos⁡−1xy = \cos^{-1} x is not sinusoidal; it is a monotonically decreasing curve.

Correct Answer: True

Solution:

The secant function, sec⁡(x)\sec(x), is defined such that its range is R−(−1,1)R - (-1, 1).

Correct Answer: True

Solution:

The cosine function has a range of [-1, 1], and its graph oscillates between these values.

Correct Answer: True

Solution:

Inverse trigonometric functions are important in calculus as they are used to define many integrals.

Correct Answer: False

Solution:

The graph of the inverse cosine function is not sinusoidal; it is a curve that represents the inverse relationship.

Correct Answer: True

Solution:

The principal value of cot⁡−1x\cot^{-1} x is indeed in the interval (0,π)(0, \pi).

Correct Answer: True

Solution:

The principal value branch of cot⁡−1\cot^{-1} is (0,π)(0, \pi), and cot⁡−11=π4\cot^{-1} 1 = \frac{\pi}{4}, which lies in this interval.

Correct Answer: False

Solution:

Trigonometric functions are not naturally one-one and onto over their natural domains and ranges, which is why restrictions are needed for their inverses to exist.

Correct Answer: False

Solution:

Trigonometric functions are not one-one and onto over their natural domains and ranges, which is why their inverses do not exist without restrictions.

Correct Answer: False

Solution:

sin⁡−1x\sin^{-1}x refers to the inverse sine function, while (sin⁡x)−1(\sin x)^{-1} refers to the reciprocal of the sine function.

Correct Answer: True

Solution:

The principal value branch of the sine function is indeed [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Correct Answer: False

Solution:

The inverse cosine function, y=extcos−1(x)y = ext{cos}^{-1}(x), does not oscillate. Its range is [0,extπ][0, ext{π}], and it is a monotonically decreasing function.

Correct Answer: True

Solution:

The range of the principal value branch of cot⁡−1x\cot^{-1} x is indeed (0,π)(0, \pi). This is a standard range for the principal branch of the inverse cotangent function.

Correct Answer: True

Solution:

The range of the principal value branch of the cotangent inverse function is indeed (0, π) as stated in the excerpts.

Correct Answer: True

Solution:

The principal value is defined as the value of an inverse trigonometric function that lies within the range of its principal branch.

Correct Answer: False

Solution:

Trigonometric functions are not one-one and onto over their natural domains, which is why their inverses do not exist without restrictions.

Correct Answer: False

Solution:

The range of the principal value branch of the inverse sine function is actually [- rac{ ext{π}}{2}, rac{ ext{π}}{2}].

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Class 12 Science (PCM)

Mathematics Part 1

Inverse Trigonometric Functions

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