# NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 (Ex 2.1)

Mathematics in Class 11 and 12 can be tough, however, the benefits that it brings forth are noteworthy. Mathematics is an extremely significant subject. Therefore, it is a good idea to take up Mathematics. Mathematics opens a huge number of career opportunities for students. Even students from Humanities and Medical streams take up Mathematics to keep a secondary, safe career path. Going through the NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.1 will help the students enhance their understanding.

CBSE Class 12 Syllabus for Mathematics is quite interesting and engaging. Students are recommended to go through the syllabus of Class 12 Mathematics before starting to learn the chapters in Class 12. Mathematics requires in-depth knowledge of each topic. It is important to solve exercises given in the textbook from time to time. There are many examples given in the NCERT textbook of Class 12 Mathematics related to each topic to help students easily understand the applications of formulae and enhance their understanding of topics. To solve the Ex 2.1 Class 12th, students are required to apply exact formulas learned in Class 12 Chapter 2. Students having problems in solving 12th Maths 2.1 questions can access NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.1 from Extramarks website and mobile application.

Students need to practice effectively for a subject like Mathematics. The  NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.1 can be downloaded for boosting the board exam preparation. The latest CBSE board syllabus for Mathematics comprises six units, which are as follows:-

1.   Relations and Functions
2.   Algebra
3.   Calculus
4.   Vectors and Three-Dimensional Geometry
5.   Linear Programming
6.   Probability

Unit 1 is about Relations and Functions has two subparts to it which are:-

1.   Relations and Functions
2.   Inverse Trigonometric Functions

In Chapter 2, the primary focus is on part b i.e. Inverse Trigonometric Functions. The Chapter in question here dwells upon the study of the uses and Scopes of Trigonometric Functions that result in their Inverses, and analyse their conduct through Graphical Representations. Certain fundamental aspects of Trigonometric functions are also explained in this chapter. Even in Calculus, these Trigonometric functions pose as crucial tools as they help in defining many Integrals. The Relations and Functions unit carries good weightage in the CBSE Board Mathematics Paper. The syllabus mentions that the topics to cover under Inverse Trigonometric Functions are:-

• Definition
• Range
• Domain
• Principal Value Branch
• Graphs of Inverse Trigonometric Functions
• Elementary Properties of Inverse Trigonometric Functions

The NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.1 is based on the concept of the Inverse Trigonometric Function. The application of these methods aids students in attaining good marks in their examinations. The provided solutions match the guidelines of CBSE regarding NCERT. Students should read through the entire answer to properly understand each step and have conceptual clarity.

## NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 (Ex 2.1)

The NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.1 provided by Extramarks is curated by Extramarks experts in a tactical way so that the solutions are easy to understand and also to the point. The step-by-step explanation and the amount of explanatory time associated with each question are highly reasonable and time-efficient. Some students usually question the importance of solving the NCERT exercises and tend to overlook them. The exercise questions help solve questions in the Mathematics board exam. Students from the Central Board of Secondary Education (CBSE) can access the solutions for all classes from the Extramarks’ website and mobile application. Extramarks has solutions for the following classes-

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Inverse Trigonometric Functions can be somewhat challenging to understand, therefore, it is likely that students look for answers to grasp the theories and tactics behind each answer. Students are advised to solve the NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.1 for all clarifications.

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## Access NCERT Solutions for Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.1 can be accessed from the Extramarks’ website and mobile application in PDF format.

### Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1

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### Access Other Exercises of Class 12 Maths Chapter 2

 Chapter 2 Inverse Trigonometric Functions Exercises Exercise 2.2 20 Questions & Solutions (4 Short Answers, 16 Long Answers)

Q.1 Find the principal values of the following

${\mathrm{sin}}^{–1}\left(–\frac{1}{2}\right)$

Ans

$\begin{array}{l}\mathrm{Let} {\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{sinx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{sin}}^{–1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\mathrm{sin}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6}\end{array}$

Q.2

${\mathrm{cos}}^{–1}\left(\frac{\sqrt{3}}{2}\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }{\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=\frac{\sqrt{3}}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cos}}^{-1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]\mathrm{and}\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{\sqrt{3}}{2}.\end{array}$

Q.3

${\text{cosec}}^{–1}\left(2\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }{\mathrm{cosec}}^{-1}\left(2\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosecx}=2.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cosec}}^{-1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]-\left\{0\right\}\mathrm{and}\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)=2.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cosec}}^{-1}\left(2\right)=\frac{\mathrm{\pi }}{6}\end{array}$

Q.4

${\text{tan}}^{–1}\left(-\sqrt{3}\right)$

Ans

$\begin{array}{l}\mathrm{Let} {\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)=\mathrm{x}.\mathrm{Then},\mathrm{tanx}=-\sqrt{3}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{tan}}^{–1}\mathrm{is}\\ \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\mathrm{and}\mathrm{tan}\left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)=-\frac{\mathrm{\pi }}{3}.\end{array}$

Q.5

${\text{cos}}^{–1}\left(-\frac{1}{2}\right)$

Ans

$\begin{array}{l}\mathrm{Let} {\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cos}}^{–1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]\mathrm{and}\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}.\\ \mathrm{cosx}=-\frac{1}{2}\\ \mathrm{ }=-\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\\ \mathrm{ }=\mathrm{cos}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ \mathrm{ }=\mathrm{cos}\left(\frac{2\mathrm{\pi }}{3}\right)\\ ⇒\mathrm{x}=\frac{2\mathrm{\pi }}{3}\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)=\frac{2\mathrm{\pi }}{3}.\end{array}$

Q.6

${\text{tan}}^{–1}\left(-1\right)$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }{\mathrm{tan}}^{-1}\left(-1\right)=-1.\\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{range} \mathrm{of} \mathrm{the} \mathrm{principle} \mathrm{value} \mathrm{branch} \mathrm{of} {\mathrm{tan}}^{-1} \mathrm{is} \\ \left(-\frac{\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2}\right) \mathrm{and} \mathrm{tan}\left(\frac{\mathrm{\pi }}{4}\right)\mathrm{ }=1.\\ \mathrm{Therefore},\mathrm{ }\mathrm{principle} \mathrm{value} \mathrm{of} {\mathrm{tan}}^{-1}\left(-1\right)=-\frac{\mathrm{\pi }}{4}\end{array}$

Q.7

${\text{sec}}^{–1}\left(\frac{2}{\sqrt{3}}\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }{\mathrm{sec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)=\mathrm{x}.\mathrm{Then},\mathrm{secx}=\frac{2}{\sqrt{3}}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cos}}^{-1} \mathrm{is}\\ \left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}\mathrm{and}\mathrm{sec}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{2}{\sqrt{3}}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{6}\end{array}$

Q.8

${\text{cot}}^{–1}\left(\sqrt{3}\right)$

Ans

$\begin{array}{l}\mathrm{Let} {\mathrm{cot}}^{-1}\left(\sqrt{3}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cotx}=\sqrt{3}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cot}}^{–1}\mathrm{is}\\ \left(0,\mathrm{\pi }\right)\mathrm{and}\mathrm{cot}\left(\frac{\mathrm{\pi }}{6}\right)=\sqrt{3}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cot}}^{-1}\left(\sqrt{3}\right)=\frac{\mathrm{\pi }}{6}.\end{array}$

Q.9

${\text{cos}}^{–1}\left(-\frac{1}{\sqrt{2}}\right)$

Ans

$\begin{array}{l}\mathrm{Let} {\mathrm{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=-\frac{1}{\sqrt{2}}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cos}}^{–1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]\mathrm{and}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}.\\ \mathrm{cosx}=-\frac{1}{\sqrt{2}}\\ \mathrm{ }=-\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\\ \mathrm{ }=\mathrm{cos}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)\\ \mathrm{ }=\mathrm{cos}\left(\frac{3\mathrm{\pi }}{4}\right)\\ ⇒\mathrm{x}=\frac{3\mathrm{\pi }}{4}\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)=\frac{3\mathrm{\pi }}{4}.\end{array}$

Q.10

$cose{c}^{–1}\left(-\sqrt{2}\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }{\mathrm{cosec}}^{-1}\left(-\sqrt{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosecx}=-\sqrt{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}\\ {\mathrm{cosec}}^{-1}\mathrm{is}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]-\left\{0\right\}\mathrm{and}\mathrm{cosec}\left(\frac{\mathrm{\pi }}{4}\right)=\sqrt{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cosec}}^{-1}\left(-\sqrt{2}\right)=-\frac{\mathrm{\pi }}{4}\end{array}$

Q.11 Find the values of the following:

${\text{tan}}^{–1}\left(1\right)+{\mathrm{cos}}^{–1}-\left(\frac{1}{2}\right)+{\mathrm{sin}}^{–1}\left(-\frac{1}{2}\right)$

Ans

$\begin{array}{l}\mathrm{Given}:\\ {\mathrm{tan}}^{–1}\left(\mathrm{1}\right)+{\mathrm{cos}}^{–1}\mathrm{}-\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{\mathrm{sin}}^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{We}\mathrm{shall}\mathrm{find}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{of}\mathrm{each}\mathrm{term}\mathrm{as}\mathrm{given}\mathrm{below}:\\ \mathrm{Let} {\mathrm{tan}}^{-1}\left(1\right)=\mathrm{x}.\mathrm{Then},\mathrm{tanx}=1.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{tan}}^{–1}\mathrm{is}\\ \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\mathrm{and}\mathrm{}\mathrm{t}\mathrm{}\mathrm{an}\left(\frac{\mathrm{\pi }}{4}\right)=1.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{tan}}^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}.\\ \mathrm{second}\mathrm{term}:\\ \mathrm{Let} {\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cos}}^{–1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]\mathrm{and}\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}.\\ \mathrm{cosx}=-\frac{1}{2}\\ \mathrm{ }=-\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\\ \mathrm{ }=\mathrm{cos}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ \mathrm{ }=\mathrm{cos}\left(\frac{2\mathrm{\pi }}{3}\right)\\ \mathrm{and}\mathrm{third}\mathrm{term}:\\ \mathrm{Let} {\mathrm{sin}}^{-1}\mathrm{}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{sinx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{sin}}^{–1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\mathrm{sin}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6}\\ \mathrm{Then},\\ {\mathrm{tan}}^{–1}\left(\mathrm{1}\right)+{\mathrm{cos}}^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)+{\mathrm{sin}}^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{\pi }}{4}+\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\\ =\frac{3\mathrm{\pi }+8\mathrm{\pi }-2\mathrm{\pi }}{12}\\ =\frac{9\mathrm{\pi }}{12}\\ =\frac{3\mathrm{\pi }}{4}.\end{array}$

Q.12

${\text{cos}}^{–1}\left(\frac{1}{2}\right)+2{\mathrm{sin}}^{–1}\left(\frac{1}{2}\right)$

Ans

$\begin{array}{l}{\mathrm{cos}}^{–1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{}+2{\mathrm{sin}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{First}\mathrm{term}:{\mathrm{cos}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{Let}\mathrm{ }{\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cos}}^{-1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]\mathrm{and}\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{-1}\mathrm{}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{3}\\ \mathrm{Second}\mathrm{term}: 2{\mathrm{sin}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{Let} {\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{sinx}=\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{sin}}^{–1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\mathrm{sin}\mathrm{}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{}\\ \mathrm{principal}\mathrm{value}\mathrm{of}2{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=2×\frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{3}\\ \therefore \mathrm{ }{\mathrm{cos}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+2{\mathrm{sin}}^{–1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{3}\\ \mathrm{ }=\frac{2\mathrm{\pi }}{3}\end{array}$

Q.13

$\begin{array}{l}{\mathrm{Ifsin}}^{–1}\mathrm{x}=\mathrm{y},\mathrm{then}\\ \\ \mathrm{A}\right)0\le \mathrm{y}\le \mathrm{\pi }\\ \mathrm{B}\right)–\frac{\mathrm{\pi }}{2}\le \mathrm{y}\le \frac{\mathrm{\pi }}{2}\\ \mathrm{C}\right)0<\mathrm{y}<\mathrm{\pi }\\ \mathrm{D}\right)–\frac{\mathrm{\pi }}{2}<\mathrm{y}<\frac{\mathrm{\pi }}{2}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},{\mathrm{sin}}^{-\mathrm{1}}:\mathrm{}\left[–\mathrm{1},1\right]\to \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\\ \mathrm{So},\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{.}\end{array}$

Q.14

$\begin{array}{l}{\mathrm{tan}}^{–1}\sqrt{3}–{\mathrm{sec}}^{–1}\left(–2\right)\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\mathrm{\pi }\\ \left(\mathrm{B}\right)–\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{C}\right)\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{D}\right)\frac{2\mathrm{\pi }}{3}\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}: {\mathrm{tan}}^{–1}\sqrt{\mathrm{3}}-{\mathrm{sec}}^{–1}\left(-\mathrm{2}\right)\\ \mathrm{Since}, \mathrm{ }{\mathrm{tan}}^{–1}\sqrt{\mathrm{3}}=\frac{\mathrm{\pi }}{3},\mathrm{and}\\ \mathrm{let} \mathrm{ }{\mathrm{sec}}^{-1}\left(-2\right)=\mathrm{x}.\mathrm{Then},\mathrm{secx}=-\mathrm{ }2.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{sec}}^{-1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}\mathrm{and}\mathrm{sec}\left(\frac{\mathrm{\pi }}{3}\right)=2.\\ \therefore \mathrm{secx}=-2=-\mathrm{sec}\left(\frac{\mathrm{\pi }}{3}\right)\\ \mathrm{ }=\mathrm{sec}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ \mathrm{ }=\mathrm{sec}\left(\frac{2\mathrm{\pi }}{3}\right)\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sec}}^{-1}\left(-2\right)=\frac{2\mathrm{\pi }}{3}.\\ \therefore \mathrm{ }{\mathrm{tan}}^{–1}\sqrt{\mathrm{3}}-{\mathrm{sec}}^{–1}\left(-\mathrm{2}\right)=\frac{\mathrm{\pi }}{3}-\left(\frac{2\mathrm{\pi }}{3}\right)\\ =\frac{\mathrm{\pi }-2\mathrm{\pi }}{3}\\ =-\frac{\mathrm{\pi }}{3}.\\ \mathrm{Option} \mathrm{B}\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}.\end{array}$