As the name suggests, this chapter's basic concept is related to Inverse Trigonometric Functions, Class 12. Sine, cosine, tangent, cot, cosec, sec are the fundamental trigonometric values that are distinctly elaborated in Chapter 2, Inverse Trigonometric Functions Class 12.

This subtopic covers these basic trigonometric functions and the ways in which you can use them to find the principal value of their inverse operations.

You will also be introduced to concepts attached to Inverse Trigonometric Functions Class 12, such as the principal value, domain and the range commonly affiliated to it. Discussing further upon the graphs associated with several inverse trigonometric functions, this chapter delves into talking briefly about the properties of all six operations.

In this subtopic of Inverse Trigonometric Functions Class 12, you will be introduced to the various properties, inter-relationships and domains of the inverse trigonometric functions.

You will also be acquainted with the calculation used to figure the operations out, their angles and certain principal values too.

Here are a few solved examples of Inverse Trigonometric Functions Class 12,

1. Find the values of cos (tan 1¾)

Answer:

Let, tan−1 3/4= θ

Therefore, tan θ = 3/4

We know that sec 2θ- tan2θ = 1

⇒ sec θ = √(1 + tan2θ)

⇒ sec θ = √(1 + (3/4)2)

⇒ sec θ = √(1 + 9/16)

⇒ sec θ = √(25/16)

⇒ sec θ = 5/4

Therefore, cos θ = ⅘ ⇒ θ = cos−14/5

Now, cos (tan−1¾) = cos (cos−14/5) = ⅘

Therefore, cos (tan−1¾) = ⅘ 

2. Simplify csc ( arctan x )

Answer:

Let z = csc ( tan x ) and y = tan x so that z = csc y = 1 / sin y. Now, after using theorem 3 above y = tan x, it can be written as,

tan y = x with - π / 2 < y < π / 2

Now, tan2y = sin2y / cos2y = sin2y / (1 - sin2y)

Solving for the above sin y

sin y = + or - √ [ tan2y / (1 + tan2y) ] = + or - | tan y | / √ [ (1 + tan2y) ]

For - π / 2 < y ≤ 0 sin y is negative and tan y is negative too, so that | tan y | = - tan y and sin y = - ( - tan y ) / √ [ (1 + tan2y) ] = tan y / √ [ (1 + tan2y) ]

For,

0 ≤ y < π/2 sin y is positive and tan y is also positive so that | tan y | = tan y an sin y = tan y / √ [ (1 + tan2y) ]

Thus, z = csc ( tan x ) = 1 / sin y = √ [ (1 + x2) ] / x

3. Find the exact value of arctan(- 1 )

Let y = arctan(- 1 ). According to the operation,

tan y = - 1 with - π / 2 < y < π / 2

Known from the table of special angles, tan (π / 4) = 1.

Now, we know that tan(- x) = - tan x.

So, tan (-π / 4) = - 1

Now considering to compare the last statement to tan y = - 1 to obtain y = - π/4

1. Sin-1x is not particularly similar to (sinx)-1. In a nutshell, (sin x)-1 = 1/sinx and relevant operations for other trigonometric functions. You’ll reap the benefits of learning several of these standpoints while studying more on Inverse Trigonometric Functions Class 12.
2. When there is no mention of the branches of inverse trigonometric functions, then it clearly means that the principal value of that function must be used.
3. To determine a value associated with an inverse trigonometric function that lies in the range of the principal branch is called the principal value of that Inverse Trigonometric Functions (Class 12).

To achieve the best score, it is important that all your doubts are cleared and that you get sufficient practice. To this end, several in-text and miscellaneous questions have been provided. Here are the key features of the NCERT Solutions for Inverse Trigonometric Functions Class 12.

1. NCERT Solutions of class 12 maths helps students strengthen fundamental concepts related to the topic.
2. The exercises and in-text questions of Inverse Trigonometric Functions Class 12 are simplified by several subject matter specialists to facilitate learning.
3. Due to the simplification, students are able to grasp concepts better and get good scores in the exam.
4. Thanks to the solutions provided, students are able to learn better. 
5. The solutions projected in Inverse Trigonometric Functions Class 12 are briefed in more than one way, assisting to broaden the knowledge of students.

1 . How Many Questions are Present in NCERT Solutions for Class 12 Maths Chapter 2? 

Meticulously prepared by maths experts, Class 12 Inverse Trigonometry Solutions typically includes three exercises. Now, the first one consists of 12 short answers and two MCQs. Following that, the second exercise consists of 18 short answers and three MCQs. After both of these, you will be introduced to a miscellaneous exercise that comes with 14 short answers and three MCQs. Your concepts will be clarified while studying the last exercise.

2 . Is NCERT Solutions for Class 12 Maths Chapter 2 Helpful in the Board Exam Preparation?

Yes, Inverse Trigonometry Class 12 Ncert Solution Chapter 2 will undeniably help you prepare for the board exams since the problems in the book are based on the curriculum CBSE follows. Prepared by several maths specialists, the expert solutions to numerous problems will help create a basic foundation to strengthen your roots. Strictly based on the latest CBSE syllabus, the NCERT solutions will boost learning in an effective way.

3 . Can NCERT Solutions for Inverse Trigonometric Functions be Used for JEE Preparation?

Yes. Since JEE Mains is partly based on the curriculum CBSE follows, these inverse trigonometric functions class 12 solutions can help you prepare thoroughly. However, it is still important that you study materials other than NCERT as well, when and where required. But, studying NCERT can contribute to better scores in JEE Mains.

4 . What Videos Should We Refer to for CBSE Class 12 Maths?

Prior to diving into conclusions, it’s important that you study thoroughly. The videos for Class 12 Inverse Trigonometric Functions can be found anywhere on our page. In fewer words, these videos will solve your queries and propel you towards high scores.

5 . How do I get Notes for CBSE Class 12 Maths Chapter 2 Online?

You can access all the important notes of Class 12 Inverse Trigonometric Functions on this page. The solutions found on our page will boost learning with easy methods to approach a problem, and several other ways to understand the concepts associated with it. Delving into the topic, the solutions provided are convenient, easy to understand, and effective during preparations. Without any time constraints, you can now download the PDF version made by numerous mathematics specialists too.