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# NCERT Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions

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## NCERT Maths Chapter 2 Inverse Trigonometric Functions Solutions of Class 12 - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 2 Solutions PDF is available for download. You can find the official PDF on Vedantu's website and download it from there. The Inverse Trigonometric Functions Class 12 PDF complete solutions have been provided for Chapter 2 of Class 12 NCERT Maths. A learner can find anything and everything related to Class 12 Maths Chapter 2. Vedantu has a team of passionate and devoted educators who are ready to depart their knowledge and expertise in the most effective way possible. In addition to this, if a student faces any doubts concerning CH 2 Maths Class 12, he or she can go to the website and drop in their queries and download NCERT Book Solution for Class 12 Maths Chapter 2 solutions PDF. They will get back to you in case of doubts and clear that off in a very efficient manner.

Table of Content
1. NCERT Maths Chapter 2 Inverse Trigonometric Functions Solutions of Class 12 - Free PDF Download
2. Glance of NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions | Vedantu
3. Practice and Master the Concepts of Real Numbers with NCERT Solutions Ex for Class 12 Maths
4. Access NCERT Solutions for Class 12 Maths  Chapter 2 - Inverse Trigonometric Functions
4.1Exercise 2.2
4.2Miscellaneous Solutions
5. Inverse Trigonometric Functions Class 12 CBSE NCERT Solutions
5.12.2 Basic Concepts
5.22.3 Properties of Inverse Trigonometric Functions
5.3Importance of Inverse Trigonometric Functions
6. Inverse Trigonometric Functions Chapter Summary - Class 12 NCERT Solutions
7. Overview of Deleted Syllabus for CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions
8. Class 12 Maths NCERT Solutions Chapter 2 All Exercises
8.1Exercise 2.1
8.214 Questions & Solutions (14 Short Answers)
8.3Exercise 2.2
8.5Miscellaneous Exercise
8.614 Question & Solutions
9. Other Study Materials of CBSE Class 12 Maths Chapter 3
10. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance of NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions | 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.

• Notation: sin⁻¹(x) = arcsine(x) = angle whose sine is x (similarly for cosine, tangent, etc.)

• Some Important identities to remember:

• sin⁻¹(sin(x)) = x for x ∈ [-π/2, π/2] (similarly for cosine and tangent)

• sin⁻¹(-x) = -sin⁻¹(x) for x ∈ [-1, 1] (similar identities for other inverse trig functions)

• Also, this article contains chapter notes, formula and exercises link and important questions for chapter 2 - Inverse Trigonometric Functions.

• There are 3 exercises (43 fully solved questions) in class 12th maths chapter 2 Inverse Trigonometric Functions.

 Current Syllabus Exercises of Class 12 Maths Chapter 2 NCERT Solutions of Class 12 Maths Inverse Trigonometric Functions Exercise 2.1 NCERT Solutions of Class 12 Maths Inverse Trigonometric Functions Exercise 2.2
Competitive Exams after 12th Science
More Free Study Material for Inverse Trigonometric Functions
Revision notes
Important questions
Ncert books

## Practice and Master the Concepts of Real Numbers with NCERT Solutions Ex for Class 12 Maths

Exercise 2.1: In this exercise, students will learn about the inverse trigonometric functions, including their domain, range, and principal values. They will also practice finding the inverse trigonometric functions of various values and solving equations involving inverse trigonometric functions.

Exercise 2.2: This exercise focuses on the properties of inverse trigonometric functions. Students will learn about the properties of sine, cosine, tangent, cotangent, secant, and cosecant functions, and practice using these properties to evaluate inverse trigonometric functions.

## Access NCERT Solutions for Class 12 Maths  Chapter 2 - Inverse Trigonometric Functions

### Exercise 2.2

1. Using inverse trigonometric identities, prove that  $3{{\sin }^{-1}}x={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right),x\in \left[ -\frac{1}{2},\frac{1}{2} \right]$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\text{x= }\!\!\alpha\!\!\text{ }$

$\text{sin }\!\!\alpha\!\!\text{ =x}$

We know that $\text{sin3 }\!\!\theta\!\!\text{ =3sin }\!\!\theta\!\!\text{ -4si}{{\text{n}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }$

Consider

$\text{si}{{\text{n}}^{\text{-1}}}\left( \text{3x-4}{{\text{x}}^{\text{3}}} \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \text{3sin }\!\!\alpha\!\!\text{ -4si}{{\text{n}}^{\text{3}}}\text{ }\!\!\alpha\!\!\text{ } \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin3 }\!\!\alpha\!\!\text{ } \right)$

$\text{=3 }\!\!\alpha\!\!\text{ }$

$\text{=3si}{{\text{n}}^{\text{-1}}}\text{x}$

Hence it is proved that $\text{3si}{{\text{n}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\left( \text{3x-4}{{\text{x}}^{\text{3}}} \right)$

2. Using inverse trigonometric identities, prove that  $3{{\cos }^{-1}}x={{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right),x\in \left[ \frac{1}{2},1 \right]$

Ans: Assuming $\text{co}{{\text{s}}^{\text{-1}}}\text{x= }\!\!\alpha\!\!\text{ }$

$\text{cos }\!\!\alpha\!\!\text{ =x}$

We know that $\text{cos3 }\!\!\theta\!\!\text{ =4co}{{\text{s}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ -3cos }\!\!\theta\!\!\text{ }$

Consider

$\text{co}{{\text{s}}^{\text{-1}}}\left( \text{4}{{\text{x}}^{\text{3}}}\text{-3x} \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left( \text{4co}{{\text{s}}^{\text{3}}}\text{ }\!\!\alpha\!\!\text{ -3cos }\!\!\alpha\!\!\text{ } \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos3 }\!\!\alpha\!\!\text{ } \right)$

$\text{=3 }\!\!\alpha\!\!\text{ }$

$\text{=3co}{{\text{s}}^{\text{-1}}}\text{x}$

Hence it is proved that $\text{3co}{{\text{s}}^{\text{-1}}}\text{x=co}{{\text{s}}^{\text{-1}}}\left( \text{4}{{\text{x}}^{\text{3}}}\text{-3x} \right)$

3. Simplify ${{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x},x\ne 0$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}\text{-1}}{\text{x}}$

Step 1:

Assuming

$\text{x=tan }\!\!\beta\!\!\text{ }$

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\text{x}$

Step 2:

Considering

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}\text{-1}}{\text{x}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ }}\text{-1}}{\text{tan }\!\!\beta\!\!\text{ }} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{sec }\!\!\beta\!\!\text{ -1}}{\text{tan }\!\!\beta\!\!\text{ }} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1-cos }\!\!\beta\!\!\text{ }}{\text{sin }\!\!\beta\!\!\text{ }} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\beta\!\!\text{ }}{\text{2}} \right) \right)$

$\text{=}\frac{\text{ }\!\!\beta\!\!\text{ }}{\text{2}}$

$\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}\text{-1}}{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

4. Simplify $\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}} \right)\text{,x }\!\!\pi\!\!\text{ }$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}}$

$\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}}$

$\text{=ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{2si}{{\text{n}}^{\text{2}}}\left( \frac{\text{x}}{\text{2}} \right)}{\text{2co}{{\text{s}}^{\text{2}}}\left( \frac{\text{x}}{\text{2}} \right)}}$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{x}}{\text{2}} \right) \right)$

$\text{=}\frac{\text{x}}{\text{2}}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}}\text{=}\frac{\text{x}}{\text{2}}$

5. Simplify $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)\text{,0x }\!\!\pi\!\!\text{ }$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)$

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1-tan x}}{\text{1+tan x}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right) \right)$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x}$

6. Simplify $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}\text{,}\left| \text{x} \right|\text{a}$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}$

Assuming

$\text{x=asin }\!\!\alpha\!\!\text{ }$ $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{asin }\!\!\alpha\!\!\text{ }}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{a}}^{\text{2}}}\text{si}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan }\!\!\alpha\!\!\text{ } \right)$

$\text{= }\!\!\alpha\!\!\text{ }$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right)$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right)$

7. Simplify ${{\tan }^{-1}}\left( \frac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right),a>0;\frac{-a}{\sqrt{3}}\le x\le \frac{a}{\sqrt{3}}$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{2}}}\text{x-}{{\text{x}}^{\text{3}}}}{{{\text{a}}^{\text{3}}}\text{-3a}{{\text{x}}^{\text{2}}}} \right)$

Assuming

$\text{x=atan }\!\!\alpha\!\!\text{ }$

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{2}}}\text{x-}{{\text{x}}^{\text{3}}}}{{{\text{a}}^{\text{3}}}\text{-3a}{{\text{x}}^{\text{2}}}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{3}}}\text{tan }\!\!\alpha\!\!\text{ -}{{\text{a}}^{\text{3}}}\text{tan }\!\!\alpha\!\!\text{ }}{{{\text{a}}^{\text{3}}}\text{-3}{{\text{a}}^{\text{3}}}\text{ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan3 }\!\!\alpha\!\!\text{ } \right)$

$\text{=3 }\!\!\alpha\!\!\text{ }$

$\text{=3ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right)$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{2}}}\text{x-}{{\text{x}}^{\text{3}}}}{{{\text{a}}^{\text{3}}}\text{-3a}{{\text{x}}^{\text{2}}}} \right)\text{=3ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right)$

8. Evaluate $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}} \right) \right]$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}}\text{=a}$

$\text{sin a=}\frac{\text{1}}{\text{2}}$

$\text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ -----(1)

Further solving,

From equation (1)

$\text{ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}} \right) \right]$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2 }\!\!\times\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right]$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right]$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right)$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}} \right) \right]\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

9. Evaluate $\text{tan}\frac{\text{1}}{\text{2}}\left[ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right]\text{,}\left| \text{x} \right|\text{1,y0 and xy1}$

Ans: Assuming $\text{x=tan }\!\!\alpha\!\!\text{ }$

Consider

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}} \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2tan }\!\!\alpha\!\!\text{ }}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin2 }\!\!\alpha\!\!\text{ } \right)$

$\text{=2 }\!\!\alpha\!\!\text{ }$

$\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x}$ ----(1)

Consider $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right)$

Assuming $\text{y=tan }\!\!\beta\!\!\text{ }$

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ }}{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ }} \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos2 }\!\!\beta\!\!\text{ } \right)$

$\text{=2 }\!\!\beta\!\!\text{ }$

$\text{=2ta}{{\text{n}}^{\text{-1}}}\text{y}$ -----(2)

From equations (1) and (2),

$\text{tan}\frac{\text{1}}{\text{2}}\left[ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right]$

$\text{=tan}\frac{\text{1}}{\text{2}}\left[ \text{2ta}{{\text{n}}^{\text{-1}}}\text{x+2ta}{{\text{n}}^{\text{-1}}}\text{y} \right]$

$\text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x+ta}{{\text{n}}^{\text{-1}}}\text{y} \right)$

$\text{=}\frac{\text{x+y}}{\text{1-xy}}$

Therefore, $\text{tan}\frac{\text{1}}{\text{2}}\left[ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right]\text{=}\frac{\text{x+y}}{\text{1-xy}}$

10. Evaluate $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

Ans: We know that $\text{sin}\left( \text{ }\!\!\theta\!\!\text{ } \right)\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ - }\!\!\theta\!\!\text{ } \right)$

Consider $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\left( \text{ }\!\!\pi\!\!\text{ -}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right) \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right) \right)$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

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

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

Ans: We know that $\text{tan}\left( \text{ }\!\!\theta\!\!\text{ } \right)\text{=-tan}\left( \text{- }\!\!\theta\!\!\text{ } \right)$

Consider $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)$

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{-tan}\left( \text{-}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)$

$\text{=-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

12. Evaluate $\text{tan}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{+co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}} \right)$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=a}$

$\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=a}$

$\text{sin a=}\frac{\text{3}}{\text{5}}$

$\text{cos a=}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}}$

$\text{cos a=}\frac{\text{4}}{\text{5}}$

$\text{tan a=}\frac{\text{3}}{\text{4}}$

$\text{a=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$ ----(1)

$\text{co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{2}}{\text{3}}$ -----(2)

Further solving,

$\text{tan}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{+co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}} \right)$

$\text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{4}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{2}}{\text{3}} \right)$

$\text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{2}}{\text{3}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{3}} \right)} \right)$

$\text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{17}}{\text{6}} \right)$

$\text{=}\frac{\text{17}}{\text{6}}$

Therefore, $\text{tan}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{+co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}} \right)\text{=}\frac{\text{17}}{\text{6}}$

13. Value of $\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$ is

(A) $\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

(B) $\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

(C) $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

(D) $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Ans: Consider $\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{-7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left[ \text{cos}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}} \right]$

$\text{=}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

Therefore, $\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

The correct option is B

14. Value of $\text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) \right)$ is

(A) $\frac{\text{1}}{\text{2}}$

(B) $\frac{\text{1}}{\text{3}}$

(C) $\frac{\text{1}}{\text{4}}$

(D) $\text{1}$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=a}$

$\text{sin a=-}\frac{\text{1}}{\text{2}}$

$\text{=-sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=sin}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Further solving,

$\text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) \right)$

$\text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{+}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$

$\text{=1}$

Therefore, $\text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) \right)\text{=1}$

The correct option is D

### Miscellaneous Solutions

1. Evaluate $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

Ans: Consider $\text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{=cos}\left( \text{2 }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=cos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

Further solving,

$\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right)$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Therefore, $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

2. Evaluate $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

Ans: Consider $\text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

Further solving,

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right)$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

3. Using inverse trigonometric identities, prove that  $\text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{24}}{\text{7}}$

Ans: Assuming $\text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }$ ----(1)

$\text{sin}\frac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\frac{\text{3}}{\text{5}}$

$\text{cos}\frac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}}$

$\text{cos}\frac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\frac{\text{4}}{\text{5}}$

Hence,

$\text{sin }\!\!\alpha\!\!\text{ =2}\left( \frac{\text{3}}{\text{5}} \right)\left( \frac{\text{4}}{\text{5}} \right)$

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{24}}{\text{25}}$

$\text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{24}}{\text{25}} \right)}^{\text{2}}}}$

$\text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{7}}{\text{25}}$

Therefore,

$\text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{24}}{\text{7}}$

$\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{24}}{\text{7}} \right)$

From Equation (1), it is proved that

$\text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{24}}{\text{7}} \right)$

4. Using inverse trigonometric identities, prove that  $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{77}}{\text{36}}$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{8}}{\text{17}}\text{= }\!\!\alpha\!\!\text{ }$

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{8}}{\text{17}}$

$\text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{8}}{\text{17}} \right)}^{\text{2}}}}$

$\text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{15}}{\text{17}}$

Hence,

Therefore,

$\text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{8}}{\text{15}}$

$\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{15}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{17}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{15}} \right)$ ----(1)

Assuming that $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }$

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}}$

$\text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}}$

$\text{cos }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}}$

Therefore,

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{4}}$

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$ ----(2)

Consider

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)$

From Equations (1) and (2),

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{15}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{8}}{\text{15}}\text{+}\frac{\text{3}}{\text{4}}}{\text{1-}\left( \frac{\text{8}}{\text{15}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}} \right)} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{77}}{\text{36}} \right)$

Hence, it is proved that $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{77}}{\text{36}}$

5. Using inverse trigonometric identities, prove that  $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\frac{\text{33}}{\text{65}}$

Ans: Assuming $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{4}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }$

$\text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{4}}{\text{5}}$

$\text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{4}}{\text{5}} \right)}^{\text{2}}}}$

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{3}}{\text{5}}$

Therefore,

$\text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{3}}{\text{4}}$

$\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{4}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$ ----(1)

Assuming that $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{= }\!\!\beta\!\!\text{ }$

$\text{cos }\!\!\beta\!\!\text{ =}\frac{\text{12}}{\text{13}}$

$\text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{12}}{\text{13}} \right)}^{\text{2}}}}$

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{5}}{\text{13}}$

Therefore,

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{5}}{\text{12}}$

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$ ----(2)

Consider

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)$

From Equations (1) and (2),

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{33}} \right)$

$\text{=co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{33}}{\text{65}} \right)$

Hence, it is proved that $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\frac{\text{33}}{\text{65}}$

6. Using inverse trigonometric identities, prove that  $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{56}}{\text{65}}$

Ans: Assuming $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }$

$\text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{12}}{\text{13}}$

$\text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{12}}{\text{13}} \right)}^{\text{2}}}}$

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}}$

Therefore,

$\text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}}$

$\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$ ----(1)

Assuming that $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }$

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}}$

$\text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}}$

$\text{cos }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}}$

Therefore,

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{4}}$

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$ ----(2)

Consider

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)$

From Equations (1) and (2),

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{33}} \right)$

$\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{65}} \right)$

Hence, it is proved that $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{56}}{\text{65}}$

7. Using inverse trigonometric identities, prove that  $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }$

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}}$

$\text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{5}}{\text{13}} \right)}^{\text{2}}}}$

$\text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{12}}{\text{13}}$

Therefore,

$\text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}}$

$\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)$ ----(1)

Assuming that $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }$

$\text{cos }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}}$

$\text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}}$

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}}$

Therefore,

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{3}}$

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right)$

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right)$ ----(2)

Consider

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)$

From Equations (1) and (2),

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{4}}{\text{3}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{63}}{\text{16}} \right)$

Hence, it is proved that $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}$

8. Using inverse trigonometric identities, prove that  ${{\tan }^{-1}}\sqrt{x}=\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-x}{1+x} \right),x\in \left[ 0,1 \right]$

Ans: Assuming $\text{x=ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }$

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}$

$\text{=ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}$

$\text{=ta}{{\text{n}}^{\text{-1}}}\text{tan }\!\!\alpha\!\!\text{ }$

$\text{= }\!\!\alpha\!\!\text{ }$

$\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{= }\!\!\alpha\!\!\text{ }$----(1)

Consider

$\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)$

$\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right)$

$\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{cos2 }\!\!\alpha\!\!\text{ }$

$\text{= }\!\!\alpha\!\!\text{ }$

$\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\text{= }\!\!\alpha\!\!\text{ }$ -----(2)

From Equations (1) and (2),

It is proved that $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)$

9. Using inverse trigonometric identities, prove that  ${{\cot }^{-1}}\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\frac{x}{2},x\in \left( 0,\frac{\pi }{4} \right)$

Ans: We know that

$\sqrt{\text{1+sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}}$

$\sqrt{\text{1-sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}}$

Consider

$\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right)$

$\text{=co}{{\text{t}}^{\text{-1}}}\left( \frac{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{+}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)}{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{-}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)} \right)$

$\text{co}{{\text{t}}^{\text{-1}}}\left( \text{cot}\frac{\text{x}}{\text{2}} \right)$

$\text{=}\frac{\text{x}}{\text{2}}$

It is proved that $\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right)\text{=}\frac{\text{x}}{\text{2}}$

10. Using inverse trigonometric identities, prove that  ${{\tan }^{-1}}\left( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right)=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x,-\frac{1}{\sqrt{2}}\le x\le 1$

Ans: Assuming

$\text{x=cos2 }\!\!\beta\!\!\text{ }$

$\sqrt{\text{1+x}}$

$\text{=}\sqrt{\text{1+cos2 }\!\!\beta\!\!\text{ }}$

$\text{=}\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ }$

$\sqrt{\text{1-x}}$

$\text{=}\sqrt{\text{1-cos2 }\!\!\beta\!\!\text{ }}$

$\text{=}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }$

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ -}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }}{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ +}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1-tan }\!\!\beta\!\!\text{ }}{\text{1+tan }\!\!\beta\!\!\text{ }} \right)$

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ } \right) \right)$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ }$

$\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x}$

It is proved that $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x}$

11. For what value of $x$ does the equation $\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)$ satisfy?

Ans: Consider $\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)$

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)$

$\frac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2cosecx}$

$\text{sin xcos x=si}{{\text{n}}^{\text{2}}}\text{x}$

$\text{sin x}\left( \text{cos x-sin x} \right)\text{=0}$

$\text{sin x=0,cos x-sin x=0}$

$\text{x=0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

Therefore, $\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)$ is satisfied for $\text{x=0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

12. For what value of $x$ does the equation $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1-x}}{\text{1+x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x,}\left( \text{x > 0} \right)$ satisfy?

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1-x}}{\text{1+x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x} \right) \right)\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

$\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

$\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}\frac{\text{3}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

$\text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{x=}\frac{\text{1}}{\sqrt{\text{3}}}$

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1-x}}{\text{1+x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$ is satisfied for $\text{x=}\frac{\text{1}}{\sqrt{\text{3}}}$

13. The expression $\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{,}\left| \text{x} \right| < \text{1}$ is equal to

(A) $\frac{\text{x}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}$

(B) $\frac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}$

(C) $\frac{\text{1}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

(D) $\frac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

Ans: Assuming

$\text{x=tan }\!\!\beta\!\!\text{ }$

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\text{x}$

$\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)$

$\text{=sin }\!\!\beta\!\!\text{ }$

$\text{=}\frac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

Hence, $\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{=}\frac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

Therefore, the correct option is option D

14. For what value of $x$ does the equation $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ satisfy?

(A) $\text{0,}\frac{\text{1}}{\text{2}}$

(B) $\text{1,}\frac{\text{1}}{\text{2}}$

(C) $0$

(D) $\frac{\text{1}}{\text{2}}$

Ans: Consider

$\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)$

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)$ -----(1)

Assuming

$\text{co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\text{= }\!\!\beta\!\!\text{ }$

$\text{cos }\!\!\beta\!\!\text{ =1-x}$

$\text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \text{1-x} \right)}^{\text{2}}}}$

$\text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}$

$\text{ }\!\!\beta\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}$

$\text{co}{{\text{s}}^{\text{-1}}}\text{1-x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}$ -----(2)

Substituting Equation (2) in (1)

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}$

$\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)\text{=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}$

$\text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{=}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}$

$\text{4}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{4}}}\text{=2x-}{{\text{x}}^{\text{2}}}$

$\text{4}{{\text{x}}^{\text{4}}}\text{-5}{{\text{x}}^{\text{2}}}\text{+2x=0}$

$\text{x=0,}\frac{\text{1}}{\text{2}}\text{,}\frac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{17}}}{\text{4}}$

But considering the given options and also when $\text{x=}\frac{\text{1}}{\text{2}}$, the equation $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ doesn’t satisfy

Thus, $\text{x=0}$ is the only solution.

Hence the correct option is C

## Inverse Trigonometric Functions Class 12 CBSE NCERT Solutions

2.1 Introduction: In NCERT Class 12th Maths Chapter 2, students will delve deeper into the concepts of Inverse Trigonometric Functions Class 12 NCERT Solutions. One will have to assess what they learnt in the previous chapter of relations and functions and the trigonometric functions studied in Class 11. Therefore, the assessment will pave the way for a deeper understanding of the foundational base of Class 12 Maths Chapter 2 NCERT solutions. Students will learn the restrictions on domains and ranges of trigonometric functions that ensure their inverses and observe their behaviour through graphical representations. A student needs to revise the previous sections of trigonometric functions to understand Inverse trigonometric functions class 12 NCERT solutions. Keeping in mind that trigonometric functions are not one-one and onto over their natural domains and ranges and their inverses do not exist, you can go forward.

Other than this, a refined understanding of Chapter 2 can be found by looking at the provided examples. The questions will encourage a learner to think outside the box and gain a better approach to apprehending Inverse Trigonometric Functions. The level of difficulty will gradually accelerate as one keeps on solving. It will help you get a full idea of what Inverse Trigonometric Functions Class 12 Solutions are all about.

### 2.2 Basic Concepts

A student has to reflect on what they studied in Class 11 Trigonometric Functions and in Chapter 1 of Relations and Functions to get a better grasp on Chapter 2. Corresponding invertible functions from the previous chapter paves the way for obtaining graphs of functions. The concepts also make it clear that the graph of an inverse function can be extracted from the correlated graph of the original function as a mirror image. This will further pave the way for understanding the domain and the co-domain that adheres to the principles of functions.

### 2.3 Properties of Inverse Trigonometric Functions

This particular section deals with proving some essential properties of inverse trigonometry Class 12 functions. They are the inverse functions of cosine, sine, tangent, cotangent, secant and cosecant functions with restricted domains. The results are valid within the principal value branches of the corresponding Inverse Trigonometric Functions and wherever defined. The properties of Inverse Trigonometric Functions assist in the calculation of an angle. Some results may not be valid for all values of the domains of Inverse Trigonometric Functions. They will be considered justifiable only for some values of x, for which Inverse Trigonometric Functions are defined.

### Importance of Inverse Trigonometric Functions

In the fields of engineering, construction, and architecture, inverse trigonometric ratios are widely used. Inverse trigonometric ratios are the easiest way to find an unknown angle, therefore we utilize them in places where we need to know the angle for our help and quickly get the correct values.

The following are a few examples of inverse trigonometric ratios in use.

• A right-angled triangle's unknown angles are measured using this formula.

• Used to determine the angle of inclination or depression.

• It is used by architects to measure the angle of a bridge and its supports.

• Carpenters use this to achieve a particular cut angle.

## Inverse Trigonometric Functions Chapter Summary - Class 12 NCERT Solutions

• The inverse trigonometric functions are the inverse functions of the trigonometric functions.

• The domain and the range of the trigonometric functions are converted into the range and domain of the inverse trigonometric functions.

• The domains and ranges (principal value branches) of inverse trigonometric function are given in the following table:

 Functions Domain Range (Pincipal value Branches) $y = \sin^{-1}x$ [-1, 1] $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ $y = \cos^{-1}x$ [-1, 1] $\left[0, \pi\right]$ $y = \text{cosec}^{-1}x$ $\mathbb{R}$ - (-1, 1) $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - {0}$ $y = \sec^{-1}x$ $\mathbb{R}$ - (-1, 1) $\left[0,\pi\right] - \dfrac{\pi}{2}$ $y = \tan^{-1}x$ $\mathbb{R}$ $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ $y = \cot^{-1}x$ $\mathbb{R}$ $\left(0, \pi\right)$

• $\sin ^{-1} x$ should not be confused with $(\sin x)^{-1}$. In fact $(\sin x)^{-1}-\frac{1}{\sin x}$ and similarly for other trigonometric functions,

• The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions.

• For suitable values of domain, we have

 $y = \sin^{-1}x \Rightarrow x = \sin y$ $x= \sin y \Rightarrow y = \sin^{-1} x$ $\sin(\sin^{-1}x) = x$ $\sin^{-1}(\sin x) = x$ $\sin^{-1}\dfrac{1}{x}=\text{cosec}^{-1}x, \cos^{-1}\dfrac{1}{x}=\sec^{-1}x, \tan^{-1}\dfrac{1}{x}=\cot^{-1}x$ $\sin^{-1}(-x) = -\sin^{-1}x$ $\sin^{-1}(-x) = \pi - -\cos^{-1}x$ $\tan^{-1}(-x) = -\tan^{-1}x$ $\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}x$ $\cot^{-1}(-x) = \pi - \cot^{-1}x$ $\sec^{-1}(-x) = \pi - \sec^{-1}x$ $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}, \tan^{-1}x+\cot^{-1}x = \dfrac{\pi}{2}, \text{cosec}^{-1}x+\sec^{-1}x = \dfrac{\pi}{2}$ $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\dfrac{x+y}{1-xy}, \tan^{-1}x+\tan^{-1}y = \pi+\tan^{-1}\dfrac{x+y}{1-xy}, xy > 1, x,y > 0$ $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\dfrac{x-y}{1+xy}$ $2\tan^{-1}xy= \tan^{-1}\dfrac{2x}{1-x^2}$ $2\tan^{-1}x = \sin^{-1}\dfrac{2x}{1+x^2} = cos^-1{1-x^2}{1+x^2}$

## Overview of Deleted Syllabus for CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions

 Chapter Dropped Topics Inverse Trigonometric Functions 2.3 Properties of Inverse Trigonometric Functions (Except sin (sin-1 x) = x, x ∈ [-1, 1] sin-1 (sin x) = x, x ∈ [-π/2, π/2] Page Number 45 – 47 Examples 4, 7 and 8 Alternative Solution of Example 5 Page Number 47- 48:Questions - 3, 4, 6, 12, 14, 15 Page Number 49-51Examples 10, 11, 12, 13 Page 51-52 Miscellaneous ExerciseQuestion number - 8, 12, 17 Page 53 Summary Points from 8 to 13

## Class 12 Maths NCERT Solutions Chapter 2 All Exercises

Chapter 2 - Inverse Trigonometric Functions All Exercises in PDF Format

## Conclusion

The NCERT Solutions for Class 12 Maths on inverse trigonometric functions provided by Vedantu are crucial for understanding this complex topic in a simple manner. It's important to focus on the concept of inverse trigonometric functions, their properties, and their applications in real-life scenarios. These solutions help students grasp the fundamental concepts and solve problems with ease. Previous year question papers typically include around 3-4 questions related to inverse trigonometric functions. Vedantu's NCERT Solutions offer a comprehensive understanding of inverse trigonometric functions, aiding students in scoring well and building a strong foundation in mathematics.

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions

1. How many questions are there in NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions?

The NCERT Class 12 Chapter 2 is based on the Inverse Trigonometric Functions. There are a total of 3 exercises in this chapter. There are 14 sums in the first exercise (Ex.-2.1) of NCERT Solutions for Inverse Trigonometric Functions. There are 20 sums in the second exercise Ex-2.2. Followed by these two exercises, there is a miscellaneous exercise in this chapter. The miscellaneous exercise consists of 17 sums that cover all the concepts of Inverse Trigonometric Functions that are explained in this chapter.

2. What is meant by inverse trigonometric functions?

The inverse trigonometric functions help to determine the angle values for the given trigonometric ratios. These functions commonly include sin-1 x, cos-1 x, tan-1 x, cot-1 x, cosec-1 x, sec-1 x. There are various formulas that relate these inverse trigonometric functions.

It is very important to learn and practice all these formulas so as to understand their applications by solving the sums given in the practice exercises of NCERT Class 12 Maths Chapter 2 Inverse Trigonometric Functions.

3. Where can I get relevant NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions online?

You can get one of the most relevant NCERT Solutions for Chapter 2 Maths Class 12 Inverse Trigonometric Functions on Vedantu. These NCERT Solutions are among the top-rated study resources for the Class 12 topic of Inverse Trigonometric Functions. You can download the PDF comprising these solutions for free. Also, NCERT Solutions for this chapter are available on our mobile application. These solutions are prepared by the subject matter experts at Vedantu in strict adherence to the NCERT guidelines for Class 12. So, students can rely on these NCERT solutions for Class 12 Maths Chapter 2 for their exam preparation.

4. Can I refer to the NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions for my IIT JEE preparation?

Yes, you can refer to the NCERT Solutions for Chapter 2 Maths Class 12 Inverse Trigonometric Functions for your IIT JEE preparation as well. These NCERT solutions are worked out in a step by step method that helps students to understand the logical analysis of every sum of Inverse Trigonometric Functions. When students develop a deeper understanding of the problem-solving techniques for every type of sum covered in this chapter, only then they can solve these sums much faster during their IIT JEE examination.

5. What is the domain and range of inverse trigonometric functions?

Although the domain of inverse trigonometric functions varies depending on the particular function, they typically have narrow domains to guarantee one-to-one correspondence.

For Example, sin^-1(x) has a domain of [-1, 1] and its range is [−π/2,π/2].

6. Why do we restrict inverse trig functions?

We restrict inverse trigonometric functions to make sure they give us only one answer for each input number. The functions wouldn't function correctly without these limitations/restrictions since they wouldn't be "one-to-one," which means that there would be several possible output values for each input integer. By limiting or restricting the input numbers(domain) and output responses(range), we ensure that each input has a unique outcome, which makes functions easier to understand and use.

7. What are the conditions for inverse trigonometric functions?

Restricted Domain: Inverse trigonometric functions have restricted domains to ensure they are one-to-one functions. This indicates that there is only one output value that corresponds to each input value.

Principal Value Ranges: The primary range of values for each inverse trigonometric function indicates the range of probable solutions it can provide. These value sets are carefully chosen to provide consistency

8.  What is the limit of inverse trigonometry class 12 NCERT Solutions?

The limits of inverse trigonometric functions depend on which specific function you're considering and the direction (positive or negative infinity) from which you're approaching the limit.

For Tangent:

• lim (as x approaches positive infinity) of arctan(x) = π/2

• lim (as x approaches negative infinity) of arctan(x) = -π/2

For Sine:

• lim(as x approaches positive infinity) of arcsin(x)=π/2

• lim(as x approaches negative infinity) of arcsin(x)=-π/2

For Cosine:

• lim(as x approaches positive infinity) of arccos(x)=0

• lim(as x approaches negative infinity) of arcsin(x)=π

9. What is the application of inverse function?

Here are some real time examples of inverse functions:

• As the speed of the car increases the time taken to cover a certain distance decreases.

• More buses on the road means less space on the road.

• The number of people doing something and the time it takes to do it. As the number of people increases, the time it takes to finish decreases.