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Properties of Inverse Trigonometric Functions

Q: What is Properties of Inverse Trigonometric Functions in Class 12 Maths?

Properties of Inverse Trigonometric Functions is part of Chapter 2 (Inverse Trigonometric Functions) in NCERT Class 12 Mathematics. This topic is fundamental for CBSE board exams and competitive tests.

🎓 Class 12 Mathematics CBSE Theory Ch 2 — Inverse Trigonometric Functions ⏱ ~25 min

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2.3 Properties of Inverse Trigonometric Functions

In this section, we establish some important properties of inverse trigonometric functions. These results are valid within the principal value branches^? of the corresponding functions and wherever they are defined.

Property Set 1: Cancellation Properties

Self-Inverse Properties

\(\sin(\sin^{-1} x) = x\), for \(x \in [-1, 1]\) and \(\sin^{-1}(\sin x) = x\), for \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
\(\cos(\cos^{-1} x) = x\), for \(x \in [-1, 1]\) and \(\cos^{-1}(\cos x) = x\), for \(x \in [0, \pi]\)
\(\tan(\tan^{-1} x) = x\), for \(x \in \mathbf{R}\) and \(\tan^{-1}(\tan x) = x\), for \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Similar results hold for \(\cot^{-1}\), \(\sec^{-1}\), and \(\csc^{-1}\).

Property Set 2: Negative Argument Identities

Odd/Even Behaviour

\(\sin^{-1}(-x) = -\sin^{-1}(x)\), for \(x \in [-1, 1]\) (odd function)
\(\tan^{-1}(-x) = -\tan^{-1}(x)\), for \(x \in \mathbf{R}\) (odd function)
\(\csc^{-1}(-x) = -\csc^{-1}(x)\), for \(|x| \geq 1\) (odd function)
\(\cos^{-1}(-x) = \pi - \cos^{-1}(x)\), for \(x \in [-1, 1]\)
\(\sec^{-1}(-x) = \pi - \sec^{-1}(x)\), for \(|x| \geq 1\)
\(\cot^{-1}(-x) = \pi - \cot^{-1}(x)\), for \(x \in \mathbf{R}\)

Memory Aid

Odd functions (sin, tan, csc): Their inverses satisfy \(f^{-1}(-x) = -f^{-1}(x)\) (the minus sign passes through).

Even-like functions (cos, sec, cot): Their inverses satisfy \(f^{-1}(-x) = \pi - f^{-1}(x)\) (supplement rule).

Property Set 3: Complementary Pair Identities

Sum = Constant

\(\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\), for \(x \in [-1, 1]\)
\(\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}\), for \(x \in \mathbf{R}\)
\(\csc^{-1}(x) + \sec^{-1}(x) = \frac{\pi}{2}\), for \(|x| \geq 1\)

Property Set 4: Reciprocal Identities

Reciprocal Relations

\(\sin^{-1}\left(\frac{1}{x}\right) = \csc^{-1}(x)\), for \(|x| \geq 1\)
\(\cos^{-1}\left(\frac{1}{x}\right) = \sec^{-1}(x)\), for \(|x| \geq 1\)
\(\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}(x)\), for \(x > 0\)

Property Set 5: Addition Formulas for tan¹

tan¹ Addition

\[\tan^{-1} x + \tan^{-1} y = \begin{cases}\tan^{-1}\left(\dfrac{x + y}{1 - xy}\right), & \text{if } xy < 1 \\[8pt] \pi + \tan^{-1}\left(\dfrac{x + y}{1 - xy}\right), & \text{if } xy > 1,\; x > 0 \\[8pt] -\pi + \tan^{-1}\left(\dfrac{x + y}{1 - xy}\right), & \text{if } xy > 1,\; x < 0\end{cases}\]
Subtraction form: \[\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x - y}{1 + xy}\right), \quad xy > -1\]

Property Set 6: Double Angle Formulas

2sin¹, 2cos¹, 2tan¹

\(2\sin^{-1} x = \sin^{-1}(2x\sqrt{1-x^2})\), for \(-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\)
\(2\cos^{-1} x = \cos^{-1}(2x^2 - 1)\), for \(0 \leq x \leq 1\)
\(2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\), for \(-1 < x < 1\)
\(2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right)\), for \(-1 \leq x \leq 1\)
\(2\tan^{-1} x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\), for \(x \geq 0\)

Worked Example 3

Show that:

(i) \(\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1} x\), for \(-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\)

(ii) \(\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\cos^{-1} x\), for \(\frac{1}{\sqrt{2}} \leq x \leq 1\)

Solution

(i) Let \(x = \sin \theta\). Then \(\sin^{-1} x = \theta\). We have:
\(2x\sqrt{1 - x^2} = 2\sin\theta\sqrt{1 - \sin^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta\)

So \(\sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1} x\).
(Valid when \(-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}\), i.e., \(-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\).)

(ii) Let \(x = \cos \theta\). Then proceeding similarly, we get \(2\cos^{-1} x\).

Worked Example 4

Express \(\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right)\), \(-\frac{3\pi}{2} < x < \frac{\pi}{2}\), in the simplest form.

Solution

We write:
\[\frac{\cos x}{1 - \sin x} = \frac{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2} = \frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2}\] \[= \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\] Therefore, \(\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right) = \frac{\pi}{4} + \frac{x}{2}\).

Worked Example 5

Write \(\cot^{-1}\left(\frac{1}{\sqrt{x^2 - 1}}\right)\), \(x > 1\) in the simplest form.

Solution

Let \(x = \sec \theta\), then \(\sqrt{x^2 - 1} = \tan \theta\).

\(\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right) = \cot^{-1}(\cot \theta) = \theta = \sec^{-1} x\), which is the simplest form.

Exercise 2.2

Prove the following:

Q1. \(3\sin^{-1} x = \sin^{-1}(3x - 4x^3)\), \(x \in \left[-\frac{1}{2}, \frac{1}{2}\right]\)

Let \(x = \sin \theta\), so \(\sin^{-1} x = \theta\) where \(\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]\).

RHS = \(\sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1} x\) = LHS.
(Since \(3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), the principal value branch applies.)

Q2. \(3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)\), \(x \in \left[\frac{1}{2}, 1\right]\)

Let \(x = \cos \theta\), so \(\cos^{-1} x = \theta\) where \(\theta \in \left[0, \frac{\pi}{3}\right]\).

RHS = \(\cos^{-1}(4\cos^3\theta - 3\cos\theta) = \cos^{-1}(\cos 3\theta) = 3\theta = 3\cos^{-1} x\) = LHS.
(Since \(3\theta \in [0, \pi]\), the principal value branch applies.)

Write the following functions in the simplest form:

Q3. \(\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\), \(x \neq 0\)

Let \(x = \tan \theta\). Then \(\sqrt{1+x^2} = \sec \theta\).
\(\frac{\sec\theta - 1}{\tan\theta} = \frac{1 - \cos\theta}{\sin\theta} = \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}\)

So the expression = \(\tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}\tan^{-1} x\).

Q4. \(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\), \(0 < x < \pi\)

Using half-angle identities: \(\frac{1 - \cos x}{1 + \cos x} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2}\).

So \(\sqrt{\frac{1-\cos x}{1+\cos x}} = \left|\tan\frac{x}{2}\right| = \tan\frac{x}{2}\) (since \(0 < x < \pi\) means \(0 < \frac{x}{2} < \frac{\pi}{2}\)).

Therefore, \(\tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}\).

Q5. \(\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)\), \(0 < x < \pi\)

Dividing numerator and denominator by \(\cos x\):
\(\frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right)\)

So \(\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right) = \frac{\pi}{4} - x\).

Find the values of each of the following:

Q8. \(\tan^{-1}\left(\tan\frac{3\pi}{4}\right)\)

Note: \(\frac{3\pi}{4}\) is NOT in the range \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) of \(\tan^{-1}\).

\(\tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\frac{\pi}{4} = -1\).

So \(\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}\).

Q10. \(\sin^{-1}\left(\sin\frac{2\pi}{3}\right)\)

\(\frac{2\pi}{3} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so we cannot directly cancel.
\(\sin\frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\).

\(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\) (which lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)).

Q14. If \(\sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1} x\right) = 1\), find \(x\).

\(\sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1} x\right) = 1\) means \(\sin^{-1}\frac{1}{5} + \cos^{-1} x = \frac{\pi}{2}\).

So \(\cos^{-1} x = \frac{\pi}{2} - \sin^{-1}\frac{1}{5} = \cos^{-1}\frac{1}{5}\).

(Using: \(\sin^{-1} a + \cos^{-1} a = \frac{\pi}{2}\), so \(\frac{\pi}{2} - \sin^{-1}\frac{1}{5} = \cos^{-1}\frac{1}{5}\).)

Therefore \(x = \frac{1}{5}\).

Q15. If \(\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} = \frac{\pi}{4}\), find \(x\).

Using the addition formula \(\tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)\) (when \(ab < 1\)):

Let \(a = \frac{x-1}{x-2}\), \(b = \frac{x+1}{x+2}\).

\(a + b = \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)} = \frac{x^2+x-2+x^2-x-2}{x^2-4} = \frac{2x^2-4}{x^2-4} = \frac{2(x^2-2)}{x^2-4}\)

\(1 - ab = 1 - \frac{(x-1)(x+1)}{(x-2)(x+2)} = 1 - \frac{x^2-1}{x^2-4} = \frac{x^2-4-x^2+1}{x^2-4} = \frac{-3}{x^2-4}\)

So \(\tan^{-1}\left(\frac{2(x^2-2)}{-3}\right) = \frac{\pi}{4}\).

\(\frac{2(x^2-2)}{-3} = \tan\frac{\pi}{4} = 1\), giving \(2(x^2-2) = -3\), so \(2x^2-4 = -3\), \(2x^2 = 1\), \(x^2 = \frac{1}{2}\).

\(x = \pm\frac{1}{\sqrt{2}}\).

Activity: Verifying Complementary Identities

Predict: Is it always true that \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\) regardless of x?

Choose \(x = 0\): \(\sin^{-1} 0 = 0\), \(\cos^{-1} 0 = \frac{\pi}{2}\). Sum = \(\frac{\pi}{2}\). Verified!
Choose \(x = 1\): \(\sin^{-1} 1 = \frac{\pi}{2}\), \(\cos^{-1} 1 = 0\). Sum = \(\frac{\pi}{2}\). Verified!
Choose \(x = -\frac{1}{2}\): \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\), \(\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\). Sum = \(-\frac{\pi}{6} + \frac{2\pi}{3} = \frac{-\pi + 4\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\). Verified!
Try \(x = 2\): \(\sin^{-1}(2)\) is undefined. The identity only holds for \(x \in [-1, 1]\).

Observation: The identity holds for every value of \(x\) in the common domain \([-1, 1]\), but is meaningless outside it.

Proof sketch: Let \(\sin^{-1} x = \theta\). Then \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\sin \theta = x\). Now \(\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta = x\). Since \(\frac{\pi}{2} - \theta \in [0, \pi]\) (the range of \(\cos^{-1}\)), we get \(\cos^{-1} x = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \sin^{-1} x\). Rearranging: \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\).

Interactive: Verify Inverse Trig Identities

Enter a value and verify key identities numerically

x (in [-1, 1]):

Competency-Based Questions

In a physics experiment, a student studies projectile motion. The angle of launch \(\alpha\) can be computed from the formula \(\alpha = \frac{1}{2}\sin^{-1}\left(\frac{gR}{v^2}\right)\), where \(g = 10\) m/s², \(R\) is the range, and \(v\) is the initial speed.

Q1. If \(v = 20\) m/s and the desired range is \(R = 20\) m, find the launch angle \(\alpha\).

L3 Apply

\(\alpha = \frac{1}{2}\sin^{-1}\left(\frac{10 \times 20}{400}\right) = \frac{1}{2}\sin^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} \times \frac{\pi}{6} = \frac{\pi}{12} = 15^\circ\).

Q2. For the same speed, what range gives a launch angle of \(30^\circ\)?

L3 Apply

\(30^\circ = \frac{1}{2}\sin^{-1}\left(\frac{10R}{400}\right)\), so \(\sin^{-1}\left(\frac{R}{40}\right) = 60^\circ = \frac{\pi}{3}\).
\(\frac{R}{40} = \sin 60^\circ = \frac{\sqrt{3}}{2}\), giving \(R = 20\sqrt{3} \approx 34.64\) m.

Q3. What is the maximum possible range, and what angle achieves it?

L4 Analyse

Maximum range occurs when \(\sin^{-1}\left(\frac{gR}{v^2}\right) = \frac{\pi}{2}\), i.e., \(\frac{gR}{v^2} = 1\), giving \(R_{max} = \frac{v^2}{g} = \frac{400}{10} = 40\) m.
At this point, \(\alpha = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4} = 45^\circ\). This matches the well-known result that a \(45^\circ\) launch angle maximizes range.

Q4. Using the identity \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\), rewrite \(\alpha\) in terms of \(\cos^{-1}\).

L4 Analyse

Since \(\sin^{-1} t = \frac{\pi}{2} - \cos^{-1} t\):
\(\alpha = \frac{1}{2}\left(\frac{\pi}{2} - \cos^{-1}\left(\frac{gR}{v^2}\right)\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\left(\frac{gR}{v^2}\right)\).

Assertion–Reason Questions

Assertion (A): \(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}\)
Reason (R): \(\sin^{-1}(-x) = -\sin^{-1}(x)\) for all \(x \in [-1, 1]\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both are true. By R: \(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}\). R directly gives A.

Assertion (A): \(\cos^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{3}\)
Reason (R): \(\cos^{-1}(-x) = \pi - \cos^{-1}(x)\) for \(x \in [-1, 1]\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) — A is false. Using R: \(\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\), NOT \(-\frac{\pi}{3}\). Also, \(\cos^{-1}\) always returns values in \([0, \pi]\), so a negative output is impossible. R is true.

Assertion (A): \(\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi\)
Reason (R): \(\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\) when \(xy > 1\) and \(x, y > 0\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Using R: \(\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
Then \(\tan^{-1}(1) + \frac{3\pi}{4} = \frac{\pi}{4} + \frac{3\pi}{4} = \pi\). Both A and R are true, and R is used to prove A.

Frequently Asked Questions

What is sin-inverse(x) + cos-inverse(x)?

For all x in [-1, 1], sin-inverse(x) + cos-inverse(x) = pi/2. Similarly, tan-inverse(x) + cot-inverse(x) = pi/2 for all real x.

What are the key properties of inverse trig functions?

Key properties: sin-inverse(-x) = -sin-inverse(x), cos-inverse(-x) = pi - cos-inverse(x), tan-inverse addition formula, and double-angle conversions.

How to simplify expressions involving inverse trig functions?

Convert to a common inverse trig form, use complementary relations, apply double-angle formulas like 2tan-inverse(x) = sin-inverse(2x/(1+x^2)), and use addition formulas.

What is the formula for tan-inverse(x) + tan-inverse(y)?

tan-inverse(x) + tan-inverse(y) = tan-inverse((x+y)/(1-xy)) when xy < 1. Additional cases apply when xy > 1.

How to prove inverse trigonometric identities?

Let the expression equal theta, use the direct trig value, apply known identities, convert all terms to one type using standard conversions, and apply addition formulas.

Frequently Asked Questions — Inverse Trigonometric Functions

What is Properties of Inverse Trigonometric Functions in NCERT Class 12 Mathematics?

Properties of Inverse Trigonometric Functions is a key concept covered in NCERT Class 12 Mathematics, Chapter 2: Inverse Trigonometric Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Properties of Inverse Trigonometric Functions step by step?

To solve problems on Properties of Inverse Trigonometric Functions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 2: Inverse Trigonometric Functions?

The essential formulas of Chapter 2 (Inverse Trigonometric Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Properties of Inverse Trigonometric Functions important for the Class 12 board exam?

Properties of Inverse Trigonometric Functions is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Properties of Inverse Trigonometric Functions?

Common mistakes in Properties of Inverse Trigonometric Functions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Properties of Inverse Trigonometric Functions?

End-of-chapter NCERT exercises for Properties of Inverse Trigonometric Functions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.

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