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Class 12-science RD SHARMA Solutions Maths Chapter 4 - Inverse Trigonometric Functions

Inverse Trigonometric Functions Exercise Ex. 4.1

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

Inverse Trigonometric Functions Exercise Ex. 4.2

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)

Inverse Trigonometric Functions Exercise Ex. 4.3

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)

Inverse Trigonometric Functions Exercise Ex. 4.4

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(i)

Solution 2(ii)

Solution 3(i)

Solution 3(ii)

Inverse Trigonometric Functions Exercise Ex. 4.5

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)

Inverse Trigonometric Functions Exercise Ex. 4.6

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)

Inverse Trigonometric Functions Exercise Ex. 4.7

Solution 1(i)

We know that,

  

  

Hence,

Solution 2(ii)

We know that,

  

  

Hence,   

Solution 2(iv)

We know that,

  

  

Hence,

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)

Inverse Trigonometric Functions Exercise Ex. 4.8

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

Inverse Trigonometric Functions Exercise Ex. 4.9

Solution 1(iii)

  

Hence,   

Solution 1(i)

Solution 1(ii)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3

Inverse Trigonometric Functions Exercise Ex. 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

Given:

  

Hence,   

Inverse Trigonometric Functions Exercise Ex. 4.11

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 2

W e space k n o w space t h a t comma space tan to the power of minus 1 end exponent A minus tan to the power of minus 1 end exponent B equals tan to the power of minus 1 end exponent open parentheses fraction numerator A minus B over denominator 1 plus A B end fraction close parentheses comma space i f space A B greater than minus 1
C o n s i d e r space t h e space g i v e n space e x p r e s s i o n space tan to the power of minus 1 end exponent open parentheses x over y close parentheses minus tan to the power of minus 1 end exponent open parentheses fraction numerator x minus y over denominator x plus y end fraction close parentheses :
tan to the power of minus 1 end exponent open parentheses x over y close parentheses minus tan to the power of minus 1 end exponent open parentheses fraction numerator x minus y over denominator x plus y end fraction close parentheses equals tan to the power of minus 1 end exponent open parentheses fraction numerator x over y minus fraction numerator x minus y over denominator x plus y end fraction over denominator 1 plus open parentheses x over y close parentheses open parentheses fraction numerator x minus y over denominator x plus y end fraction close parentheses end fraction close parentheses
equals tan to the power of minus 1 end exponent open parentheses fraction numerator x open parentheses x plus y close parentheses minus y open parentheses x minus y close parentheses over denominator y open parentheses x plus y close parentheses plus x open parentheses x minus y close parentheses end fraction close parentheses
equals tan to the power of minus 1 end exponent open parentheses fraction numerator x squared plus y squared over denominator x squared plus y squared end fraction close parentheses
equals tan to the power of minus 1 end exponent open parentheses 1 close parentheses
equals straight pi over 4


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 3(x)

Given:

As

  

Solution 3(xi)

Given:

As

  

But   does not satisfy the given equation.

Therefore,   

Solution 4

 

Inverse Trigonometric Functions Exercise Ex. 4.12

Solution 1

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Given:   

  

Inverse Trigonometric Functions Exercise Ex. 4.13

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Let   

  

Let

  

We know that, sin(x + y) = sinx cosy + cosx siny

  

  

Inverse Trigonometric Functions Exercise Ex. 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 4(iii)

Let

  

  

Hence,

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 x equals n straight pi plus straight pi over 4

Solution 8(v)

Given:

  

Solution 8(vi)

Given:

Using the formula

  

Solution 9

Solution 10

Solution 11

Inverse Trigonometric Functions Exercise 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 36

Given:    (i)

We know that

  

As we know that

  

Comparing the above inequality with (i), we get

  

Solution 37

Let

  

We know that,

  

Solution 38

Let

  

We know that,

  

Solution 39

Given:   

We know that

  

Solution 40

One branch of cos-1 other than the principal value branch corresponds to  .

Solution 41

The principal value branch of sec-1is  .

Solution 42

The principal value branch of tan-1is  .

Solution 43

The principal value branch of cosec-1is  .

Solution 44

Let

  

We know that

  

Solution 45

Given: 3tan-1x + cot-1x = π

  

Solution 46

  

Hence,

Solution 47

We know that,

  when xy < 1

Therefore,   is valid when

  

Hence, the given expression is valid for

Solution 48

We know that,

 , which is the principal value branch of tan-1x

Now,   

As

  

Inverse Trigonometric Functions Exercise Ex. 4VSAQ

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

  

Hence,

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Let

  

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

We know that,

  

  

Hence,

Solution 47

Let

  

The range of   is R - (-1, 1).

So,   does not belong to this range.

Thus,   is undefined.

Hence,

Solution 48

  

  

  

Hence,

Solution 49

Using the property that,   when   we have

  

Solution 50

We know that   when   

As   we have

  

Solution 51

  

Hence,   

Solution 52

Let

  

Now, range of cosec x is R - (-1, 1)

As   does not belong to this range, so cosec x also does not belong to this range.

Thus, the value of x will be undefined.

Hence,

Solution 53

For x < 0,

  

Hence,   

Solution 54

  for all

Solution 55

We know that,

  

Solution 56

Given:

  

Solution 57

  

Hence,   

Solution 58

Given:

  

Hence, the value of x is

Solution 59

  

Hence,   

Solution 60

  

Hence,   

Solution 61

  

As

  

Hence,

Solution 62

  

As

  

Hence,

Solution 63

Let

  

Now,   

  

Let   be the third angle.

  

Hence, the third angle is

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