CBSE Class 12 Maths Inverse Trigonometric Functions NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Maths chapter on Inverse Trigonometric Functions explores their properties and applications. The NCERT Solutions offer comprehensive guidance on evaluating expressions and proving identities related to these functions. Students will grasp essential concepts like the principal value range and fundamental properties such as $\tan^{-1}(\tan x) = x$ and $\cos^{-1}(\cos x) = x$, provided the arguments fall within the defined principal value intervals. The solutions meticulously guide learners through simplifying intricate expressions, effectively employing trigonometric identities, and applying the core definitions of inverse trigonometric functions. This resource is crafted to foster a deeper understanding and boost problem-solving skills, serving as an excellent tool for both exam preparation and thorough revision.

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Quick info

BoardCBSE
ClassClass 12
SubjectMaths Exemplar
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 2

Chapter summary

This chapter focuses on Inverse Trigonometric Functions, covering their evaluation and properties. The NCERT Solutions provide step-by-step guidance for problems involving the principal values of inverse trigonometric functions and the application of identities. Students will learn to simplify expressions like $\tan^{-1}(\tan x)$ and $\cos^{-1}(\cos x)$ by considering the domain and range, and prove trigonometric identities using these functions. The solutions aim to clarify the underlying principles and methods required to solve problems effectively.

Learning outcomes

  • Understand and apply the principal value ranges of inverse trigonometric functions.
  • Evaluate expressions involving compositions of trigonometric and inverse trigonometric functions.
  • Utilize properties of inverse trigonometric functions to simplify expressions.
  • Prove identities involving inverse trigonometric functions.
  • Solve problems requiring the conversion between inverse trigonometric functions (e.g., $\cot^{-1}$ to $\tan^{-1}$).
  • Apply the relationship between inverse trigonometric functions and their corresponding angles.

Topics covered

Paper topics

  • Inverse Trigonometric Functions
  • Principal Values
  • Properties of Inverse Trigonometric Functions
  • Evaluation of Expressions
  • Trigonometric Identities
  • Domain and Range
  • tan inverse tan x
  • cos inverse cos x
  • cot inverse
  • tan inverse
  • Inverse trigonometric function identities
  • Proving identities

Important topics

  • Principal value ranges of inverse trigonometric functions
  • Evaluating $\tan^{-1}(\tan x)$ and $\cos^{-1}(\cos x)$
  • Properties: $\tan^{-1}(-x)$, $\cos^{-1}(-x)$
  • Identity: $2\tan^{-1}x$
  • Proving identities involving inverse trigonometric functions

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Questions and Solutions

Q. 1

Find the value of $\tan^{-1} \left( \tan \frac{5\pi}{6} \right) + \cos^{-1} \left( \cos \frac{13\pi}{6} \right)$.
Solution:

We need to evaluate the expression $\tan^{-1} \left( \tan \frac{5\pi}{6} \right) + \cos^{-1} \left( \cos \frac{13\pi}{6} \right)$.

We use the properties: $\tan^{-1}(\tan x) = x$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$.

For the first term, $\frac{5\pi}{6}$ is not in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. We rewrite $\tan \frac{5\pi}{6}$ using the identity $\tan(\pi - \theta) = -\tan \theta$.

\tan \frac{5\pi}{6} = \tan \left( \pi - \frac{\pi}{6} \right) = -\tan \frac{\pi}{6}

Therefore, $\tan^{-1} \left( \tan \frac{5\pi}{6} \right) = \tan^{-1} \left( -\tan \frac{\pi}{6} \right)$. Using the property $\tan^{-1}(-y) = -\tan^{-1}(y)$, we get:

\tan^{-1} \left( -\tan \frac{\pi}{6} \right) = -\tan^{-1} \left( \tan \frac{\pi}{6} \right) = -\frac{\pi}{6}

For the second term, $\frac{13\pi}{6}$ is not in the interval $[0, \pi]$. We rewrite $\cos \frac{13\pi}{6}$ using the identity $\cos(2\pi + \theta) = \cos \theta$.

\cos \frac{13\pi}{6} = \cos \left( 2\pi + \frac{\pi}{6} \right) = \cos \frac{\pi}{6}

Therefore, $\cos^{-1} \left( \cos \frac{13\pi}{6} \right) = \cos^{-1} \left( \cos \frac{\pi}{6} \right)$. Since $\frac{\pi}{6}$ is in the interval $[0, \pi]$, we have:

\cos^{-1} \left( \cos \frac{\pi}{6} \right) = \frac{\pi}{6}

Now, we add the values of the two terms:

\tan^{-1} \left( \tan \frac{5\pi}{6} \right) + \cos^{-1} \left( \cos \frac{13\pi}{6} \right) = -\frac{\pi}{6} + \frac{\pi}{6} = 0

Answer: The value of the expression is 0.

Q. 2

Evaluate $\cos \left| \cos^{-1} \left( \frac{-\sqrt{3}}{2} \right) + \frac{\pi}{6} \right|$.
Solution:

We need to evaluate $\cos \left[ \cos^{-1} \left( \frac{-\sqrt{3}}{2} \right) + \frac{\pi}{6} \right]$.

First, let's find the value of $\cos^{-1} \left( \frac{-\sqrt{3}}{2} \right)$. We know that $\cos \frac{5\pi}{6} = \frac{-\sqrt{3}}{2}$. Since $\frac{5\pi}{6}$ lies in the principal value range $[0, \pi]$ for $\cos^{-1}$, we have:

\cos^{-1} \left( \frac{-\sqrt{3}}{2} \right) = \frac{5\pi}{6}

Now substitute this value back into the expression:

\cos \left[ \frac{5\pi}{6} + \frac{\pi}{6} \right]

Add the angles inside the cosine function:

\frac{5\pi}{6} + \frac{\pi}{6} = \frac{6\pi}{6} = \pi

So the expression becomes:

\cos(\pi)

The value of $\cos(\pi)$ is -1.

Answer: The value of the expression is -1.

Q. 3

Prove that $\cot \left( \frac{\pi}{4} - 2 \cot^{-1} 3 \right) = 7$.
Solution:

We need to prove that $\cot \left( \frac{\pi}{4} - 2 \cot^{-1} 3 \right) = 7$.

Let's work with the expression inside the cotangent function. We can convert $\cot^{-1} 3$ to $\tan^{-1}$ using the identity $\cot^{-1} x = \tan^{-1} \frac{1}{x}$ for $x > 0$.

\cot^{-1} 3 = \tan^{-1} \frac{1}{3}

So, $2 \cot^{-1} 3 = 2 \tan^{-1} \frac{1}{3}$.

Now, we use the identity $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$ for $|x| < 1$. Here, $x = \frac{1}{3}$, which satisfies $|x| < 1$.

2 \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2} \right) = \tan^{-1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} \right) = \tan^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} \right) = \tan^{-1} \left( \frac{2}{3} \cdot \frac{9}{8} \right) = \tan^{-1} \left( \frac{3}{4} \right)

Now substitute this back into the original expression:

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

We can rewrite $\frac{\pi}{4}$ as $\tan^{-1}(1)$. So the expression becomes:

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

Using the identity $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a-b}{1+ab} \right)$:

\tan^{-1}(1) - \tan^{-1} \left( \frac{3}{4} \right) = \tan^{-1} \left( \frac{1 - \frac{3}{4}}{1 + 1 \cdot \frac{3}{4}} \right) = \tan^{-1} \left( \frac{\frac{1}{4}}{\frac{7}{4}} \right) = \tan^{-1} \left( \frac{1}{7} \right)

So, we have shown that $\frac{\pi}{4} - 2 \cot^{-1} 3 = \tan^{-1} \left( \frac{1}{7} \right)$.

Now, we need to find the cotangent of this value:

\cot \left( \frac{\pi}{4} - 2 \cot^{-1} 3 \right) = \cot \left( \tan^{-1} \left( \frac{1}{7} \right) \right)

Since $\cot(\tan^{-1} y) = \frac{1}{y}$ for $y \neq 0$, we have:

\cot \left( \tan^{-1} \left( \frac{1}{7} \right) \right) = \frac{1}{\frac{1}{7}} = 7

Thus, we have proved that $\cot \left( \frac{\pi}{4} - 2 \cot^{-1} 3 \right) = 7$.

Answer: The identity is proved.

Common mistakes

  • Forgetting to consider the principal value range when evaluating $\tan^{-1}(\tan x)$ or $\cos^{-1}(\cos x)$.
  • Incorrectly applying formulas for inverse trigonometric functions, especially when dealing with negative arguments.
  • Errors in algebraic manipulation when simplifying expressions or proving identities.
  • Misapplying trigonometric identities within the context of inverse functions.

Revision tips

  • Memorize the principal value ranges for all six inverse trigonometric functions.
  • Practice converting angles outside the principal range back into the correct form.
  • Work through the provided solutions step-by-step to understand the logic.
  • Focus on the properties like $\tan^{-1}(-x) = -\tan^{-1}x$ and $\cos^{-1}(-x) = \pi - \cos^{-1}x$.
  • Attempt to solve problems without looking at the solution first, then verify your steps.

Practice MCQs

Q1. What is the principal value of $\tan^{-1}(\tan(5\pi/6))$?

Q2. What is the value of $\tan^{-1}(\tan(5\pi/6)) + \cos^{-1}(\cos(13\pi/6))$?

Q3. What is the value of $\cos^{-1}(-\sqrt{3}/2)$?

Q4. The expression $\cos [ \cos^{-1}(-\sqrt{3}/2) + \pi/6 ]$ simplifies to:

Q5. To prove $\cot(\pi/4 - 2\cot^{-1}3) = 7$, we can start by rewriting the equation as:

Q6. The identity $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ is used in the proof of Q.3. What is the value of $x$ for the term $2\cot^{-1}3$?

Frequently asked questions

What are Inverse Trigonometric Functions?

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions (sine, cosine, tangent, etc.). They return an angle whose trigonometric function value is given. For example, $\sin^{-1}(x)$ gives the angle whose sine is $x$.

Why is the principal value range important for inverse trigonometric functions?

Inverse trigonometric functions are defined to be single-valued. The principal value range ensures that each input value corresponds to a unique output angle, making the function well-defined.

How do you evaluate $\tan^{-1}(\tan x)$ when $x$ is outside the principal range $(-\pi/2, \pi/2)$?

You need to rewrite the angle $x$ such that the tangent of the new angle is equal to $\tan x$ and the new angle lies within the principal range. For example, $\tan(5\pi/6) = \tan(\pi - \pi/6) = -\tan(\pi/6)$, so $\tan^{-1}(\tan(5\pi/6)) = \tan^{-1}(-\tan(\pi/6)) = -\pi/6$.

How do you evaluate $\cos^{-1}(\cos x)$ when $x$ is outside the principal range $[0, \pi]$?

Similar to tangent, you rewrite the angle $x$ so that its cosine is equal to $\cos x$ and the new angle is within $[0, \pi]$. For example, $\cos(13\pi/6) = \cos(2\pi + \pi/6) = \cos(\pi/6)$, so $\cos^{-1}(\cos(13\pi/6)) = \cos^{-1}(\cos(\pi/6)) = \pi/6$.

What is the key identity used in Question 3 to prove the given statement?

Question 3 uses the identity $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ after converting $\cot^{-1}3$ to $\tan^{-1}(1/3)$.

How can these NCERT solutions help in exam preparation?

These solutions provide clear, step-by-step explanations for each problem, helping students understand the methods and concepts. They cover common types of questions and identities, building confidence for exams.

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