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# RD Sharma Solution for Class 12 Science Mathematics Chapter 4 - Inverse Trigonometric Functions

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 12 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 4 - Inverse Trigonometric Functions.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 12 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 12 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 2(i)

Solution 2(ii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 4

Solution 5

Solution 1

Solution 2

Solution 3

Solution 4(i)

Solution 4(ii)

Solution 4(iii)

Solution 4(iv)

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

?

Solution 5(iv)

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(i)

Solution 2(ii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(i)

Solution 2(ii)

Solution 3(i)

Solution 3(ii)

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 1(vii)

Solution 1(viii)

Solution 1(ix)

Solution 1(x)

Solution 2(i)

Solution 2(iii)

Solution 2(v)

Solution 2(vi)

Solution 2(vii)

Solution 2(viii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Solution 3(vi)

Solution 3(vii)

Solution 3(viii)

Solution 4(i)

Solution 4(ii)

Solution 4(iii)

Solution 4(iv)

Solution 4(v)

Solution 4(vi)

Solution 4(vii)

Solution 4(viii)

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 5(iv)

Solution 5(v)

Solution 5(vi)

Solution 6(i)

Solution 6(ii)

Solution 6(iii)

Solution 6(iv)

Solution 6(v)

Solution 6(vi)

Solution 7(i)

Solution 7(ii)

Solution 7(iii)

Solution 7(iv)

Solution 7(v)

Solution 7(vi)

Solution 7(vii)

Solution 7(viii)

Solution 7(ix)

Solution 7(x)

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 1(vii)

Solution 1(viii)

Solution 1(ix)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 3

Solution 4

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3

## RD Sharma Solution for Class 12 Science Mathematics Chapter 4 - Inverse Trigonometric Functions Page/Excercise 4.10

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 2

[ ∏/2 - sin-1 x ] + [ ∏/2 - sin-1y ] = ∏/4

sin-1x + sin-1y = ∏ - ∏/4

sin-1x + sin-1y = 3∏/4

Solution 3

On adding both the equation

Π/2 + sin-1y - cos-1y = Π/2

[ Π/2- cos-1y ] - cos-1y = 0

cos-1y = Π/4

y = 1/2

on putting y=1/2 in 2nd equation

cos-1x - Π/4 = Π/6

cos-1x = Π/4 + Π/6

x = cos(Π/4 + Π/6)

x = cos(Π/4)cos(Π/6)-sin(Π/4)sin(Π/6)

x = (3-1)/22

Solution 4

cot(z) = 0 means z = Π/2, 3Π/2, 5Π/2 ………..

cos-1(3/5) + sin-1x = nΠ + Π/2

sin-1x = nΠ + Π/2 - cos-1(3/5)

sin-1x = nΠ + sin-1(3/5)

x = sin(nΠ + sin-1(3/5)) = (-1)n sin (sin-1(3/5))

x = (-1)n 3/5

Solution 5

[ Π/2 - cos-1x ]2 + (cos-1x)2 =17Π2/36

Π2/4 - Πcos-1x + 2(cos-1x)2 =17Π2/36

Let, cos-1x=u

2u2 - Πu + Π2/4 - 17Π2/36 = 0

2u2 - Πu - 2Π2/9 = 0

18u2 - 9Πu -2Π2 = 0

On factorizing

18u2 - 12Πu + 3Πu -2Π2 = 0

6u( 3u -2Π ) + Π( 3u -2Π ) = 0

( 3u -2Π )(6u + Π) = 0

u = -Π/6, 2Π/3

i.e. cos-1x = -Π/6, 2Π/3

but range of cos-1x is [0, π]

x = cos(Π/2 + Π/6)

x = -1/2

Solution 6

sin-1(1/5) + [ Π/2 - sin-1x ] = sin-11

sin-1(1/5) + Π/2 - sin-1x = Π/2

sin-1(1/5) - sin-1x = 0

x = 1/5

Solution 7

Π/2 - cos-1x = Π/6 + cos-1x

Π/3 = 2cos-1x

cos-1x = Π/6

x = √3/2

Solution 8

4sin-1x+cos-1x=Π

3sin-1x+sin-1x+cos-1x=Π

3sin-1x=Π/2 [sin-1x+cos-1x=Π/2]

sin-1x=Π/6

x = sinΠ/6=0.5

Solution 9

tan-1x+cot-1x=Π/2 so the above equation reduces to

cot-1x =2Π/3-Π/2 =Π/6

x= cotΠ/6 =√3

Solution 10

2tan-1x+3(Π/2)=2Π

2tan-1x=2Π-3Π/2=Π/3

tan-1x=Π/6

x=tanΠ/6=1/√3

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Solution 3(vi)

Solution 3(vii)

Solution 3(viii)

Solution 3(ix)

Solution 4

Solution 1

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3(i)

Solution 3(ii)

Solution 1

Solution 2

Solution 3

Solution 4

## RD Sharma Solution for Class 12 Science Mathematics Chapter 4 - Inverse Trigonometric Functions Page/Excercise 4.14

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 2(vi)

Solution 2(vii)

Solution 2(viii)

Solution 2(ix)

Solution 2(x)

Solution 3

Solution 4(i)

Solution 4(ii)

Solution 5

Solution 6

Solution 7(i)

Solution 7(ii)

Solution 8(i)

Solution 8(ii)

Solution 8(iii)

Solution 8(iv)

Thus, the solution is

Solution 9

Solution 10

Solution 11

## RD Sharma Solution for Class 12 Science Mathematics Chapter 4 - Inverse Trigonometric Functions Page/Excercise MCQ

Solution 1

Correct option: (a)

Solution 2

Correct option: (d)

Solution 3

Correct option: (c)

Solution 4

Correct option: (a)

Solution 5

Correct option: (a)

Solution 6

Correct option: (b)

Solution 7

Correct option: (a)

Solution 8

Correct option: (a)

Solution 9

Correct option: (a)

Solution 10

Correct option: (c)

Solution 11

Correct option: (b)

Solution 12

Correct option: (a)

Solution 13

Correct option: (c)

Solution 14

Correct option: (a)

Solution 15

Correct option: (b)

Solution 16

Correct option: (d)

Solution 17

Correct option: (c)

Solution 18

Correct option: (b)

Solution 19

Correct option: (b)

Solution 20

Correct option: (d)

Solution 21

Correct option: (d)

Solution 22

Correct option: (a)

Solution 23

Correct option: (a)

Solution 24

Correct option: (c)

Solution 25

Correct option: (d)

Solution 26

Correct option: (a)

Solution 27

Correct option: (b)

Solution 28

Correct option: (c)

Solution 29

Correct option:(a)

Solution 30

Correct option: (c)

Solution 31

Correct option: (d)

Solution 32

Correct option: (c)

Solution 33

Correct option: (a)

Solution 34

Correct option: (c)

Solution 35

Correct option: (a)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

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TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 4 - Inverse Trigonometric Functions for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 4 - Inverse Trigonometric Functions.

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CBSE XII Science - Mathematics

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