NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 2: Inverse Trigonometric Functions, Part III

NCERT Solutions for CBSE Class 12th Mathematics, Chapter 2: Inverse Trigonometric Functions are available here. Here, you will get solutions of exercise 2.2 (from question number 1 to question number 4). The questions given in this exercise are based on basic properties of inverse trigonometric functions. These questions are important CBSE Class 12 Maths board exam and other competitive exams like UPSEE, JEE Main, WBJEE etc.

NCERT Solutions for CBSE Class 12th Maths, Chapter 2: Inverse Trigonometric Functions (Exercise 2.2) are given below

Question 1: 3 sin‒1x = sin‒1 (3x ‒ 4x3), x ϵ [‒1/2, 1/2]

Solution 1:

Let x = sin ϕ

Solving RHS

sin‒1(3x ‒ 4x3) = sin‒1[3sin ϕ ‒ 4 sin3 ϕ] = sin‒1(sin3 ϕ) = 3 ϕ.

NCERT Exemplar Class 12 Maths – Chapter 2

Question 2: 3 cos‒1x = cos‒1 (4x3 ‒ 3x), x ϵ [1/2, 1]

Solution 2:

Let x = cos ϕ

Then, ϕ = cos‒1 x

RHS = cos‒1[4x3 ‒ 3x]

Question 3: tan‒1(2/11) + tan‒1(7/24) = tan‒1(1/2)

Solution 3:

LHS = tan‒1(2/11) + tan‒1(7/24)

As, tan‒1 x + tan‒1 y = tan‒1 [(x + y)/1 ‒ xy)]

Putting x = 2/11 and y = 7/24 and solving, we have

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tan‒1(2/11) + tan‒1(7/24) = tan‒1 (1/2) = RHS.

Question 4: 2 tan‒1 (1/2) + tan‒1(1/7) = tan‒1(31/17)

Solution 4:

As, 2tan‒1x = tan‒1 [2x/(1 ‒ x2)

Using above formula, we can write 2 tan‒1 (1/2) = tan‒1 (4/3)

Now, 2 tan‒1 (1/2) + tan‒1(1/7) = tan‒1 (4/3) + tan‒1(1/7)

We know that,

tan‒1 x + tan‒1 y = tan‒1 [(x + y)/1 ‒ xy)]

Putting x = 4/3 and y = 1/7 in above formula and solving we have, 

tan‒1 (4/3) + tan‒1(1/7) = tan‒1(31/17).

Download NCERT Solutions for Class 12 Mathematics Chapter 2 - Inverse Trigonometric Functions in PDF Format

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