NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions – Exercises

The topic of Inverse Trigonometric Functions is a crucial part of Class 12 Mathematics, and understanding it lays the foundation for advanced studies in calculus, physics, and engineering. Chapter 2 in the NCERT Mathematics textbook deals with this important concept, covering a range of exercises designed to help students master inverse trigonometric functions.

This article will explore the key concepts, exercises, and solutions for Chapter 2: Inverse Trigonometric Functions. We will also offer a detailed breakdown of the exercise problems, providing clarity and insight into the solutions.

Overview of Chapter 2: Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the basic trigonometric functions (sine, cosine, tangent, etc.). While trigonometric functions map angles to ratios, inverse trigonometric functions do the reverse—they map ratios back to angles. This is vital in solving many complex mathematical problems.

Definition of Inverse Trigonometric Functions

If we have a trigonometric function such as sin(x), its inverse is denoted by $\sin^{-1}(x)$, which is read as "inverse sine of x" or "arcsine of x." Similarly, the inverses of the other basic trigonometric functions are:

• $\sin^{-1}(x)$ (arcsine)
• $\cos^{-1}(x)$ (arccosine)
• $\tan^{-1}(x)$ (arctangent)
• $\cot^{-1}(x)$ (arccotangent)
• $\sec^{-1}(x)$ (arcsecant)
• $\csc^{-1}(x)$ (arccosecant)

The primary goal of inverse trigonometric functions is to find the angle corresponding to a given trigonometric ratio. For example, if you know that $\sin(\theta) = \frac{1}{2}$, the inverse sine function helps you find the angle $\theta$.

Range of Inverse Trigonometric Functions

The inverse trigonometric functions are defined over specific intervals, known as their principal values. These are:

• $\sin^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
• $\cos^{-1}(x) \in [0, \pi]$
• $\tan^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
• $\cot^{-1}(x) \in [0, \pi]$
• $\sec^{-1}(x) \in [0, \pi], x \neq (0, \frac{\pi}{2})$
• $\csc^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}], x \neq 0$

These ranges are essential to avoid ambiguity, as trigonometric functions are periodic, and a single trigonometric value corresponds to multiple angles. Limiting the range allows for a unique solution.

Importance of Inverse Trigonometric Functions in Class 12 Maths

Inverse trigonometric functions form the base for several important concepts in calculus, particularly integration. Many problems in physics, especially those dealing with waves, oscillations, and circular motion, involve inverse trigonometric functions. Therefore, mastering the content of Chapter 2 is essential for students aiming for excellence in mathematics and related fields.

Key Topics Covered in NCERT Chapter 2

1. Introduction to Inverse Trigonometric Functions
   The chapter begins with a review of trigonometric functions and their inverses, followed by an explanation of principal values and domains.

2. Properties of Inverse Trigonometric Functions
   The various properties, such as symmetry, addition formulas, and the relationship between the different inverse trigonometric functions, are discussed.

3. Simplification of Expressions
   Simplification of expressions involving inverse trigonometric functions using standard identities is a major focus in this chapter. For example:

   $$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$ $$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$$

4. Solving Equations Involving Inverse Trigonometric Functions
   Students are required to solve equations where inverse trigonometric functions are present, often combining them with algebraic and trigonometric identities.

5. Derivatives of Inverse Trigonometric Functions
   Differentiating inverse trigonometric functions is crucial for calculus. The chapter provides formulas for these derivatives, such as:

   $$\frac{d}{dx} \left( \sin^{-1} x \right) = \frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1$$ $$\frac{d}{dx} \left( \tan^{-1} x \right) = \frac{1}{1 + x^2}$$

NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions – Exercises

The exercises in Chapter 2 test the student's understanding of the properties, formulas, and applications of inverse trigonometric functions. Below is a detailed discussion of the exercises and their solutions.

Exercise 2.1: Basic Problems on Inverse Trigonometric Functions

This exercise includes basic problems to familiarize students with inverse trigonometric functions. The questions range from finding the value of expressions like $\sin^{-1} \left( \frac{1}{2} \right)$ to solving more complex identities such as:

$$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$

Sample Problem 1:

Find the value of $\sin^{-1} \left( \frac{\sqrt{3}}{2} \right)$.

Solution: We know that $\sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$. Therefore,

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

Sample Problem 2:

Find the value of $\tan^{-1} (1) + \cot^{-1} (1)$.

Solution: We know that $\tan^{-1} (1) = \frac{\pi}{4}$ and $\cot^{-1} (1) = \frac{\pi}{4}$. Therefore,

$$\tan^{-1} (1) + \cot^{-1} (1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.$$

Exercise 2.2: Advanced Problems on Inverse Trigonometric Functions

This exercise delves deeper into solving equations and simplifying expressions involving inverse trigonometric functions. It also introduces some integration concepts related to inverse functions.

Sample Problem 1:

Prove that $\sin^{-1} x + \sin^{-1} \sqrt{1 - x^2} = \frac{\pi}{2}$, for $-1 \leq x \leq 1$.

Solution: Let $\theta = \sin^{-1} x$. This implies that $\sin \theta = x$. Since $\sin^{-1} \sqrt{1 - x^2}$ is the angle whose sine is $\sqrt{1 - x^2}$, we can express this as:

$$\theta + \phi = \frac{\pi}{2},$$

where $\theta = \sin^{-1} x$ and $\phi = \sin^{-1} \sqrt{1 - x^2}$, because:

$$\sin (\theta + \phi) = \sin \left( \frac{\pi}{2} \right) = 1.$$

Thus,

$$\sin^{-1} x + \sin^{-1} \sqrt{1 - x^2} = \frac{\pi}{2}.$$

Sample Problem 2:

Simplify $2 \tan^{-1} x$, for $x > 0$.

Solution: Using the identity $\tan(2 \theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$, we get:

$$2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right), \quad \text{for} \ |x| < 1.$$

Hence, the simplified form is $\tan^{-1} \left( \frac{2x}{1 - x^2} \right)$.

Exercise 2.3: Derivatives of Inverse Trigonometric Functions

This exercise focuses on the differentiation of inverse trigonometric functions. It includes questions that require the application of the derivative formulas for inverse trigonometric functions.

Sample Problem 1:

Find $\frac{d}{dx} \left( \sin^{-1} x \right)$.

Solution: The derivative of $\sin^{-1} x$ is given by:

$$\frac{d}{dx} \left( \sin^{-1} x \right) = \frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1.$$

Sample Problem 2:

Find the derivative of $y = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right)$.

Solution: We use the chain rule for differentiation here. Let $u = \frac{x}{\sqrt{1 - x^2}}$. Then, using the derivative formula for $\tan^{-1}(u)$, we get:

$$\frac{dy}{dx} = \frac{1}{1 + \left( \frac{x}{\sqrt{1 - x^2}} \right)^2} \cdot \frac{d}{dx} \left( \frac{x}{\sqrt{1 - x^2}} \right).$$

Simplifying, we obtain the final result.

Conclusion

NCERT Chapter 2 on Inverse Trigonometric Functions plays a pivotal role in preparing students for more advanced topics in calculus and trigonometry. The exercises in this chapter help students develop a strong foundation in handling inverse trigonometric functions, solving complex equations, and applying these concepts to real-world problems. Understanding these solutions is essential for anyone looking to excel in Class 12 Mathematics and beyond.