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NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

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NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 - Free PDF Download

Exercise 2.2 in this chapter focuses on the properties, principal values, and applications of inverse trigonometric functions. This exercise is designed to help students deepen their understanding of these functions, which are essential in various fields such as calculus, physics, and engineering. NCERT for Exercise 2.2 Class 12 Maths Solutions Free PDF download and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Maths Chapter 2 Inverse Trigonometric Functions Class 12 Ex 2.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

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Glance on NCERT Solutions Maths Chapter 2 Ex 2.2 Class 12 | Vedantu

  • Inverse trigonometric functions, also called inverse sine, cosine, tangent, etc., are the inverses of the basic trigonometric functions (sine, cosine, tangent, etc.).

  • Trigonometric functions tell you what the ratio of sides is in a right triangle given an angle. Inverse trig functions do the opposite - tells you the angle measure when given the trigonometric ratio.

  • Common inverse functions include sin⁻¹(x), cos⁻¹(x), and tan⁻¹(x).

  • The principal value of an inverse trigonometric function is the specific value within a defined range.

    • For $\sin^{-1}(x)$ , the principal value range is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

    • For $\cos^{-1}(x)$ , the principal value range is [0, \pi].

    • For $\tan^{-1}(x)$ , the principal value range is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

  • Exercise 2.2 Maths Class 12 NCERT Solutions has overall 15 fully solved Questions 

  • Class 12 Maths Chapter 2 Exercise 2.2 likely involved applying these formulas to various problems.


Topics Covered in the NCERT Solutions for Class 12 Maths Chapter 2

  1. Properties  of Inverse Trigonometric Functions

    • 1. sin⁻¹(sin(x)) = x for x in [−π/2, π/2]

    • 2. cos⁻¹(cos(x)) = x for x in [0, π]

    • 3. tan⁻¹(tan(x)) = x for x in [−π/2, π/2]

    • 4. sin(sin⁻¹(x)) = x for x in [-1, 1]

    • 5. cos(cos⁻¹(x)) = x for x in [-1, 1]

    • 6. tan(tan⁻¹(x)) = x for all x

Competitive Exams after 12th Science
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Access NCERT Solutions for Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

Refer to page 10 for Exercise 2.2 in the PDF. 

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

Ans: To Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$, where $x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Let $x = \sin \theta .$ Then, ${\sin ^{ - 1}}x = \theta $.

We have given that,

R.H.S ${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) = {\sin ^{ - 1}}\left( {3\sin \theta  - 4{{\sin }^3}\theta } \right)$

$ = {\sin ^{ - 1}}(\sin 3\theta )$

$ = 3\theta $

$ = 3{\sin ^{ - 1}}x = L.H.S$

Hence proved.

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

Ans: To Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Let $x = \cos \theta $. Then, ${\cos ^{ - 1}}x = \theta $

We have given that,

R.H S =${\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

$ = {\cos ^{ - 1}}\left( {4{{\cos }^3}\theta  - 3\cos \theta } \right)$

$ = {\cos ^{ - 1}}(\cos 3\theta )$

$ = 3\theta $

$ = 3{\cos ^{ - 1}}x = L.H.S$

Hence proved.


3. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x},x \ne 0$

Ans: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x}$

By putting $x = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}x$

$\therefore {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x} = {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {{\tan }^2}\theta }  - 1}}{{\tan \theta }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sec \theta  - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{\theta }{2}} \right) = \dfrac{\theta }{2} = \dfrac{1}{2}{\tan ^{ - 1}}x$

4. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

Ans: Given that,${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$

$ = \dfrac{x}{2}$

5. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right),0 < x < \pi $

Ans: We have given that, ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)$

$ = {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(\tan x)\quad \left[ {\because \quad \dfrac{{ - y}}{{x - xy}} = {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right]$

$ = \dfrac{\pi }{4} - x$

6. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }},|x| < a$

Ans: We have given that,${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

Let’s consider, $x = a\sin \theta  \Rightarrow \dfrac{x}{a} = \sin \theta  \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

$\therefore {\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\cos \theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan \theta ) = \theta  = {\sin ^{ - 1}}\dfrac{x}{a}$

7. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\dfrac{{ - a}}{{\sqrt 3 }} \leqslant x \leqslant \dfrac{a}{{\sqrt 3 }}$

Ans: Consider, ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

Let’s consider, $x = a\tan \theta  \Rightarrow \dfrac{x}{a} = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^2} \cdot a\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3a \cdot {a^2}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^3}\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan 3\theta )$

$ = 3\theta $

$ = 3{\tan ^{ - 1}}\dfrac{x}{a}$

8. Find the value of ${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{1}{2} = x$

Then $\sin x = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$

$\therefore {\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6}$

$\therefore {\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

$ = {\tan ^{ - 1}}\left[ {2\cos \left( {2 \times \dfrac{\pi }{6}} \right)} \right]$

$ = {\tan ^{ - 1}}\left[ {2\cos \dfrac{\pi }{3}} \right]$

$ = {\tan ^{ - 1}}\left[ {2 \times \dfrac{1}{2}} \right]$

$ = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$


9. Find the value of $\tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right],|x| < 1,y > 0$ and $xy < 1$

Ans: Let consider, $x = \tan \theta $.

Then, $\theta  = {\tan ^{ - 1}}x$.

$\therefore {\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\sin ^{ - 1}}(\sin 2\theta )$

$ = 2\theta $

$ = 2{\tan ^{ - 1}}x$

Let’s assume, $y = \tan \theta .$ Then, $\theta  = {\tan ^{ - 1}}y$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)$

$ = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\cos ^{ - 1}}(\cos 2\theta )$

$ = 2\theta  = 2{\tan ^{ - 1}}y$

$\therefore \tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)} \right]$

$ = \tan \dfrac{1}{2}\left[ {2{{\tan }^{ - 1}}x + 2{{\tan }^{ - 1}}y} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right]$

$ = \dfrac{{x + y}}{{1 - xy}}$


10. Find the values of ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

As we know that ${\sin ^{ - 1}}(\sin x) = x$

If $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, which is the principal value branch of ${\sin ^{ - 1}}x$.

Here, $\dfrac{{2\pi }}{3} \notin \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

Now, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$ can be written as:

${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

$ = {\sin ^{ - 1}}\left[ {\sin \left( {\pi  - \dfrac{{2\pi }}{3}} \right)} \right]$

$ = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{3}} \right)$, where $\dfrac{\pi }{3} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

$\therefore {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left[ {\sin \dfrac{\pi }{3}} \right] = \dfrac{\pi }{3}$

11. Find the values of ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

Ans: Let’s Consider, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

As we know that ${\tan ^{ - 1}}(\tan x) = x$

If $x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, which is the principal value branch of ${\tan ^{ - 1}}x$. Here, $\dfrac{{3\pi }}{4} \notin \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.

Now, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$ can be written as:

${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ { - \tan \left( {\dfrac{{ - 3\pi }}{4}} \right)} \right] = {\tan ^{ - 1}}\left[ { - \tan \left( {\pi  - \dfrac{\pi }{4}} \right)} \right]$

${\tan ^{ - 1}}\left( { - \tan \dfrac{\pi }{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right]$ where $ - \dfrac{\pi }{4} \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

$\therefore {\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right] = \dfrac{{ - \pi }}{4}$

12. Find the values of $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{3}{5} = x$.

Then, $\sin x = \dfrac{3}{5}$

$ \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x}  = \dfrac{4}{5}$

$ \Rightarrow \sec x = \dfrac{5}{4}$

$\therefore \tan x = \sqrt {{{\sec }^2}x - 1}  = \sqrt {\dfrac{{25}}{{16}} - 1}  = \dfrac{3}{4}$

$\therefore x = {\tan ^{ - 1}}\dfrac{3}{4}$

$\therefore {\sin ^{ - 1}}\dfrac{3}{5} = {\tan ^{ - 1}}\dfrac{3}{4}\quad  \ldots (i)$

Therefore, $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{2}{3}} \right)\quad $ [Using (i) and (ii)]

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}}} \right)} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{9 + 8}}{{12 - 6}}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{17}}{6}} \right) = \dfrac{{17}}{6}$

13. Find the values of ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ is equal to

(A) $\dfrac{{7\pi }}{6}$

(B) $\dfrac{{5\pi }}{6}$

(C) $\dfrac{\pi }{3}$

(D) $\dfrac{\pi }{6}$

Ans: We know that ${\cos ^{ - 1}}(\cos x) = x$ if $x \in [0,\pi ]$, which is the principal value branch of ${\cos ^{ - 1}}x$. Here, $\dfrac{{7\pi }}{6} \notin [0,\pi ]$.

Now, ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ can be written as:

${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{ - 7\pi }}{6}} \right) = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi  - \dfrac{{7\pi }}{6}} \right)} \right]\quad [\because $

$ + x) = \cos x]$

$\therefore {\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{6}} \right) = \dfrac{{5\pi }}{6}$

The correct answer is ${\text{B}}$.

14. Find the values of $\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right)$ is equal to

(A) $\dfrac{1}{2}$

(B) $\dfrac{1}{3}$

(C) $\dfrac{1}{4}$

(D) 1

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = x$.

Then, $\sin x = \dfrac{{ - 1}}{2} =  - \sin \dfrac{\pi }{6} = \sin \left( {\dfrac{{ - \pi }}{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$. ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{\pi }{6}$

$\therefore \sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right) = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{{3\pi }}{6}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$

The correct answer is ${\text{D}}$.

15. $\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

(B) $-\dfrac{\pi}{2}$

(C) 0

(D) $2 \sqrt{3}$

Ans:

Suppose that,

$A=\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$

As we know that,

$\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$

So,

$\cot ^{-1}(-\sqrt{3})=\pi-\cot ^{-1}(\sqrt{3})$

Put the value in the equation (i),

$A=\tan ^{-1}(\sqrt{3})-\pi+\cot ^{-1}(\sqrt{3})$

As we know that,

$\cot ^{-1} x=\dfrac{\pi}{2}-\tan ^{-1} x$

So,

$\cot ^{-1}(\sqrt{3})=\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3})$

Put the value in the equation (ii),

$A=\tan ^{-1}(\sqrt{3})-\pi+\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3}) $

$A=-\dfrac{\pi}{2}$

Hence, the correct option is B.


Conclusion

NCERT Solutions for Maths Chapter 2 Inverse Trigonometric Functions, Ex2.2 class 12 by Vedantu provides clear explanations and step-by-step solutions for understanding the properties and applications of inverse trigonometric functions. This exercise emphasizes the importance of knowing the principal value ranges for inverse functions, as well as solving equations involving these functions. Students should focus on mastering these ranges and practicing how to simplify and solve related equations accurately. Previous exam papers typically include 2-3 questions from this topic, highlighting its significance in understanding the broader subject of trigonometry. Vedantu’s solutions are designed to build a strong foundation, ensuring students can handle these concepts confidently in exams.


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FAQs on NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

1. What is the correct stepwise approach to solving Class 12 Maths Chapter 2 NCERT questions on inverse trigonometric functions as per CBSE guidelines?

To solve NCERT problems on inverse trigonometric functions according to CBSE rules, first express all trigonometric expressions in their inverse form, use the principal value branch for each function, show every calculation step-by-step without omitting working, and clearly indicate the domain and range. The final answer should be highlighted as per the official answer structure.

2. Why is it important to use the principal value branch while answering inverse trigonometric questions for NCERT Solutions?

Using the principal value branch ensures that each inverse trigonometric function produces a unique and well-defined answer within a specified interval. CBSE marking focuses on correct branch usage because it avoids ambiguity and aligns with the NCERT syllabus standards. Marks might be deducted if answers fall outside the principal value range.

3. How can common mistakes in writing Class 12 NCERT Solutions for inverse trigonometric functions be avoided?

  • Avoid skipping steps in transformations—show each derivation clearly.
  • Always justify the choice of principal value branch for each function.
  • Refrain from using answers outside the accepted domain and range.
  • Check that every solution matches the NCERT structure for CBSE boards.
  • Box or highlight the final answer as shown in the NCERT exemplar and solution book.

4. In what ways do NCERT Solutions for Class 12 Maths Chapter 2 help students prepare for CBSE 2025–26 exams?

NCERT Solutions provide structured, stepwise answers following the official CBSE pattern. This helps students understand the logic behind each step, practice clear solution writing, and prepare for typical question framing and marking used in board exams. They also expose students to a variety of question types, ensuring board exam readiness.

5. How does one solve inequalities involving inverse trigonometric functions in Class 12 Maths accurately according to NCERT Solutions?

  • First, rewrite the inequality using the relevant inverse trigonometric functions.
  • Apply the principal value ranges for each function to determine valid solution sets.
  • Justify each transformation step.
  • Clearly state the domain, range, and final form of the inequality, as per the CBSE/NCERT answer format.

6. What approach should be taken to ensure the answer format matches the CBSE 2025–26 expectations for Maths Chapter 2 solutions?

Use stepwise explanation for each question, highlight the principal value, justify domain and range, and structure the answer using the official NCERT language and formatting. Always place the boxed or underlined final answer at the conclusion of the solution.

7. How do solving Class 12 NCERT Solutions for inverse trigonometric functions benefit higher-order thinking questions?

By practicing stepwise answers, students develop the ability to logically connect statements, justify transformations, and handle complex or HOTS (higher-order thinking skills) questions in exams. This practice helps break down challenging problems into simpler steps, which is essential for scoring in board exams and competitive tests.

8. What is the significance of accurately stating domains and ranges in NCERT Solutions for Exercise 2.2?

Accurately specifying domain and range ensures the answer is mathematically sound and adheres to the rules of inverse trigonometric functions. CBSE examiners allocate marks for proper domain and range justification, and incomplete or missing domains can result in loss of marks.

9. What misconceptions might students have when solving inverse trigonometric function problems, and how can NCERT Solutions resolve them?

  • Misinterpreting principal value ranges—NCERT Solutions clarify the correct intervals with worked examples.
  • Forgetting to express results within specified domains—solutions explicitly mention domains.
  • Assuming all inverse functions are defined everywhere—solutions point out when functions are undefined.

10. Are NCERT Solutions for Class 12 Maths Chapter 2 valid for both English and Hindi medium students appearing for CBSE 2025–26?

Yes, NCERT Solutions for Class 12 Maths Chapter 2 are available in both English and Hindi, strictly designed per the NCERT textbooks and CBSE 2025–26 syllabus for all mediums. Each solution uses the official language and format required by CBSE.

11. How can students use NCERT Solutions for Class 12 Maths to build a strong foundation in inverse trigonometric functions?

By following methodical explanations and carefully written steps in NCERT Solutions, students can understand both the concept and proper answer presentation. This strengthens their conceptual base and builds confidence for trigonometry-based problems in boards and entrance exams.

12. What higher-order concept skills are developed by consistently practicing Class 12 NCERT inverse trigonometric solutions?

  • Logical reasoning for transformations and simplifications.
  • Clear understanding of trigonometric identities and their inverses.
  • Ability to solve complex equations using systematic steps.
  • Skill in recognizing and applying principal value concepts across various problem types.