Solution

PRACTICE PAPER

Class 12 - Mathematics

1.
(b) [-1, 2)
Explanation: Let f(x) = cos-1[x]
Now, domain of g(x) = cos-1x is the set
- 1 ≤ x ≤ 1} = [-1, 1]
∴ Domain of given function is {x : -1 ≤ [x] ≤ 1}

⎧ −1 if −1 ≤ x ≤ 0
⎪

[x] = ⎨ 0 if 0 ≤ x ≤ 1
⎩
⎪
1 if 1 ≤ x < 2

∴ Domain of cos-1 [x] is [-1, 2).

my
2.
(c) 1

√1+x2

Explanation: cot −1
x = θ ⇒ x = cot θ ⇒ cot θ =
x

de
P erp. 1
−1
sin(co t x) = sin θ = = .
Hyp. √x2 +1

−π
3. (a) 4

Explanation: Let the principle value be given by x

a
ac
also, let x = cosec-1 (−√2)
–

–
⇒ cosec x = − √2

–
⇒ cosec x =-cosec ( π
) (∵ cosec (
π
) = √2)
th

4 4

⇒ cosec x = cosec (− π

4
) (∵ - cosec (θ) = cosec (-θ))
⇒ x=− π
rar

4.
(b) [0, π] − { π

2
}

Explanation: To Find: The range of sec-1(x)

ast

Here, the inverse function is given by y = f-1(x)

The graph of the function y = sec-1(x) can be obtained from the graph of
Y = sec x by interchanging x and y axes.i.e, if (a, b) is a point on Y = sec x then (b, a) is the point on the function y = sec-1(x)
Sh

Below is the Graph of the range of sec-1(x)

From the graph, it is clear that the range of sec-1(x) is restricted to interval
π
[0, π] − { }
2

5. (a) 0.96
Explanation: Let sin–1 (0.6) = θ, i.e, sin θ =0.6
Now, sin (2θ) = 2 sinθ cosθ = 2 (0.6) (0.8) = 0.96
−1
6. Let cot −1
( ) = θ
√3

1/9
Er.anurag sir's classes
 −1
cot θ =
√3

We know that θ ∈ (0, π)

π
cot θ = cot(π − )
3

2π
θ=
3

−1
Therefore, principal value of cot −1
( ) =
2π

3
√3

7. The domain of sin-1x is [-1,1] and that of cos x is R. Therefore, domain of f(x) = sin-1x + cos x is [-1, 1] ∩ R = [-1, 1].
8. Let sin −1
(
1
) = θ
√2

1
⟹ sin θ =
√2

−π π
We know that θ ∈ [ 2
,
2
]

π
⟹ sin θ = sin
4
π
⟹ θ=
4

Therefore, principal value of sin −1

(
1
) is π

4
√2

9. We know that for any x ∈ R, tan-1x represents an angle in (− π

,
π
) whose tangent is x. Therefore,

my
2 2

tan-1(1) = ( An angle θ ∈ (−
π π
, ) such that tan θ = 1)
2 2
π
=
4

√3
10. sin −1
[cos(sin
−1

2
)]

de
−1 π
= sin [cos( )]
3

−1 1 π
= sin [ ] =
2 6

11. Since tan(tan-1x) = x, ∀ x ∈ R, tan (tan-1 (-4)) = - 4

12. Let cos (− ) = θ
−1 1
a
ac
2
−1
cos θ =
2

θ ∈ [0, π]

π
cos θ = cos(π − )
th

2π
θ=
3

Principal value is 2π
rar

13. Let sin-1 ( x and cos-1 (

− √3 −1

2
)=
2
) =y
− √3 −1
⇒ sin x = ( 2
) and cos y = 2

we know that the range of the principal value branch of sin-1 ( and cos-1 is (0, π)
−π
ast

π
, )
2 2

−π − √3 −1
We also know that sin 3
=( 2
) and cos (​​ 2π

3
) = 2

Value of sin-1 ( + cos-1 (

− √3 −1 −π 2π π
∴ ) ) = + =
2 2 3 3 3
Sh

√3
14. Using the principal values, we have to find the value of sin −1
(−
2
).

√3
We have, sin −1
(−
2
) = sin
−1
(− sin
π

3
)

π √3
[∵ sin = ]
3 2

−1 π
= sin [sin(− )] [∵ − sin θ = sin(−θ)]
3

π −1 π π
= − [∵ sin (sin θ) = θ; ∀θ ∈ [− , ]]
3 2 2

√3
Hence, sin −1
(−
2
) = −
π

√3 √3
15. If cos −1
(
2
) = θ , then cos θ = 2
.
√3
Since, we are considering principal branch, θ ∈ [0, π] . Also, since 2
> 0 , θ being in the first quadrant, hence
√3
cos
−1
(
2
) =
π

6
.

16. cot-1x represents an angle in (0, π ) whose cotangent is x.

–
Let x = cot (√3) −1

– π
⇒ cot x = √3 = cot( )
6

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Er.anurag sir's classes
 π
⇒ x =
6
–
∴ Principal value of cot −1
(√3) is π

6
.
17. Let sec −1
(
2
) = y
√3

2
⇒ sec y =
√3

π
⇒ sec y = sec
6

Since, the principal value branch of sec-1 is [0, π] .

Therefore, Principal value of sec −1
(
2
) is π

6
.
√3

−π
18. We have, tan −1
(−
1
) + cot
−1
(
1
) + tan
−1
[sin(
2
)] .
√3 √3

5π π
= tan
−1
(tan
6
) + cot
−1
(cot
3
) + tan
−1
(−1) .
−1 π −1 π −1 π
= tan [tan(π − )] + cot [cot( )] +tan [tan(π − )]
6 3 4

π π
−1
⎡ ∵ tan (tan x) = x, x ∈ (− , ) ⎤
2 2
−1 π −1 π −1 π ⎢ ⎥
= tan (− tan ) + cot (cot ) + tan (− tan ) ⎢ −1 ⎥
6 3 4 ⎢ cot (cot x) = x, x ∈ (0, π) ⎥
⎣ −1 −1 ⎦

my
and tan (−x) = − tan x
π π π −2π+4π−3π
= − + − =
6 3 4 12
−5π+4π −π
= =
12 12

19. We have to find principle value of , cos −1

(
1

2
)

de
Lets cos-1 1

2
= θ

1
⇒
2
= cos θ
⇒ cos
π
= cos θ
⇒ θ=
3
π a
ac
3

20. We have, tan −1

(tan
2π

3
) = tan
−1
tan(π −
π

3
)

−1 π −1 −1
= tan (− tan ) [∵ tan (−x) = − tan x]
3

−1 −π π −1 −π π
th

= tan tan ( )= − [∵ tan (tan x) = x, x ∈ ( , )]

3 3 2 2

2π 2π
Note: Remember that, tan −1
(tan
3
) ≠
3

tan-1(tan
rar

Since, x) = x, if x ∈ (− π

2
,
π

2
) and 2π

3
∉ (
−π

2
,
π

2
)

21. We know that tan −1

1 =
π

4
.
−1 −1
∴ cot[ sin {cos(tan 1)}]
ast

−1 π −1 1 π
= cot{sin (cos )} = cot(sin ) = cot = 1
4 √2 4

–
22. Let cot −1
(√3) = y
–
⇒ cot y = √3
π
Sh

⇒ cot y = cot
6

Since, the principal value branch of cot-1 is [0, π] .

–
Therefore, principal value of cot (√3) is . −1 π

6
– –
23. tan −1
√3 − sec
−1
(−2) = tan
−1
√3 − [π − sec
−1
2]

π −1 1
= − π + cos ( )
3 2
2π π π
= − + = −
3 3 3

− √3 π
24. We have, cos[ cos −1
(
2
) +
6
]

−1 π π
cos[ cos ( − cos ) + ]
6 6

−1 5π π
= cos[ cos (cos ) + ]
6 6

5π π −1
= cos( + ){∵ cos cos x = x, x ∈ [0, π]}
6 6

6π
= cos( )
6

= cos(π) = −1

25. The domain of sin-1 x is [-1,1]. Therefore, f(x) = sin-1 (-x2) is defined for all x satisfying -1 ≤ -x2 ≤ 1
2
⇒ 1 ≥ x ≥ −1

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 2
⇒ 0 ≤ x ≤ 1

2
⇒ x ≤ 1

2
⇒ x − 1 ≤ 0

⇒ (x - 1)(x + 1) ≤ 0
⇒ −1 ≤ x ≤ 1

Hence, the domain of f(x) = sin-1 (-x2) is [-1, 1].

26. Let cos −1
(−
1
) = y ,. Then, cos y = − 1
= − cos(
π

4
) = cos(π −
π

4
) = cos(
3π

4
) .
√2 √2

We know that the range of the principal value branch of cos-1 is [0, π ] and cos( 3π

4
) = −
1

√2

Therefore, the principal value of cos −1

(−
1
) is 3π

4
.
√2
–
27. Let cos ec −1
(− √2) = y
–
⇒ cos ecy = − √2
−π
⇒ cos ecy = cos ec
4

Since, the principal value branch of cosec-1 is [− π

2
,
π

2
] .
– −π
Therefore, principal value of cos ec −1
(− √2) is .

y
4

28. sin −1
(sin(− 600 )) = sin
∘ −1
{sin(−600 ×
π

180
)}

m
−1 10π −1 10π
= sin {sin(− )} = sin (− sin )
3 3

−1 π −1 π
= sin {− sin(3π + )} = sin {− (− sin )}

de
3 3

−1 π π
= sin (sin ) =
3 3

–
29. Let tan −1
(− √3) = y ca
–
⇒ tan y = − √3
π
⇒ tan y = − tan
3

π
ha
⇒ tan y = tan(− )
3

Since, the principal value branch of tan-1 is [− π

2
,
π

2
] .
–
Therefore, principal value of tan −1
(− √3) is − . π

3π 3π 3π −π
30. tan as π
art

−1
(tan ) ≠ ∉ ( , )
π 4 4 2 2

−1 3π −1 π
∵ tan (tan ) = tan [tan(π − )]
4 4

−1 π
= tan [− tan( )]
r

π
= −
ast

31. Principal value branch of tan-1 x is [− π

2
,
π

2
]

and its graph is shown below.

Sh

−π
32. From Fig. we note that tan x is an increasing function in the interval ( 2
,
π

2
) , since 1 > π

4
⇒ tan 1 > tan
π

4
. This gives
tan 1 > 1
π
⇒ tan 1 > 1 >
4

−1
⇒ tan 1 > 1 > tan (1)

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Er.anurag sir's classes
33. cosec-1x represents an angle in [− π

2
, 0) ∪ (0,
π

2
] whose cosent is x.
Let x = cosec-1(-2)
π
⇒ cosec x = −2 = cosec (− )
6

π
⇒ x = −
6

Principal value of cosec-1(-2) is − .

π
∴
6

34. Principal value branch of cos-1 x is [0, π ] and its graph is shown here,

my
35. Let cosec-1 (−√2) = y . Then, cosec y = −√2 = −cosec (
– – π

4
) = cosec (−
π

4
) .
We know that the range of the principal value branch of

de
cosec-1 is [−
–
π

2
,
π

2
] − {0} and cosec (− π

4
) = − √2 .

Therefore, the principal value of cosec-1(−√2) is − .

– π

9π
36. tan −1
(tan
8
)
a
ac
−1 π
= tan tan(π + )
8

−1 π
= tan (tan( ))
8

π
=
8
th

−1
37. Let sin −1
(
2
) = y

1
⇒ sin y = −
2
rar

π
⇒ sin y = − sin
6

π
⇒ sin y = sin(− )
6

Since, the principal value branch of sin-1 is [− π

,
π
]
ast

2 2

−1
Therefore, principal value of sin −1
(
2
) is − . π

−−−−−
38. Let y = sin-1(2x √1 − x2 )
let x = cos θ
Sh

⇒ θ = cos-1 x θ ≤ θ ≤ π

4
−−−−− −− −
⇒ y= sin-1(2cos θ √1 − sin 2 θ )
= sin-1 sin
2θ
= 2θ [θ ≤ θ ≤ π

4
⇒ θ ≤ 2θ ≤ π

2
]
⇒ 2cos-1 x
Hence Proved
39. We have, cos-1( = cos-1(cos
1 π
) )
2 3

= π

3
[∵
π

3
∈ [0, π]]

Also sin-1(− 1

2
) = sin-1 (-sin π

6
)
= sin-1 (sin(− π )) 6

=- π

6
[∵ −
π

6
∈ [−
π

2
,
π

2
]]

∴ cos-1( 1

2
) - 2sin-1(− 1

2
) = π

3
- 2(− π

6
)

= π

3
+ π

3
= 2π

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Er.anurag sir's classes
40. tan −1
(−
1
) + cot
−1
(
1
) + tan
−1
(sin(−
π

2
))
√3 √3

π π −1
= − + + tan (−1)
6 3
π π π
= − + −
6 3 4
π
= −
12
1 π
41. We know that sin −1

2
=
6
.
−1 −1 1
∴ tan {2 cos(2 sin )}
2

−1 π
= tan {2 cos(2 × )}
6

−1 π −1 1 −1 π
= tan (2 cos ) = tan (2 × ) = tan 1 =
3 2 4

π
42. Let cos −1
(
1

2
)= x . Then, cos x = 1

2
= cos(
3
) .
−1 1 π
∴ cos ( )=
2 3

Let sin −1
(
1

2
)= y . Then, sin y = 1

2
= sin(
π

6
) .
−1 1 π
∴ sin ( )=
2 6

−1 1 −1 1 π 2π π π 2π
∴ cos ( ) + 2 sin ( ) = + = + =
2 2 3 6 3 3 3

y
43. First of all we need to find the principal value for cosec–1(–2)
Let,

m
cosec–1(-2 )=Y
cosec y = –2

de
⇒

⇒ –cosec y = 2
⇒ –cosec =2 π

As we know that cosec(–θ) = –cosecθ

∴ –cosec π
= cosec (
−π
)
ca
6 6

The range of principal value of cosec–1 (-2) is [

−π π
–{0} and
ha
, ]
2 2

−π
cosec ( 6
) = –2

Thus,the princi value of cosec–1(–2) is

−π

6
.
∴ Now, the question changes to
art

Sin–1[cos
−π

6
]
Cos(–θ) = cos(θ)
r

∴ we can write the above expression as

ast

Sin–1[cos ] π

Let,
Sin–1(
√3
) =Y
2
Sh

√3
⇒ sin y = 2

⇒ Y= π

The range of principal value of sin–1 is (

−π √3

2
,
π

2
) and sin ( π

3
) =
2

Sin–1( ) is
√3
Therefore, the principal value of 2
π

π
Hence, the principal value of the given equation is 3

44. Let x = cos θ

∴ θ = cos-1 x
sec-1( 1
) = sec-1( 1
)= sec-1( 1
) = sec-1(sec 2θ)
2x2 −1 2 cos 2 θ−1 cos 2θ

= 2θ
= 2 cos-1 x
45. We know that the range of principal value of cosec-1 is [
−π

2
,
π

2
] - [0]
Let cosec-1 (-1) = θ. Then we have, cosec θ = -1
−π
cosec θ = -1 = -cosec = cosec ( ) π

2 2

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Er.anurag sir's classes
 −π −π
∴ θ=
2
∈ [
2
,
π

2
] - [0]
Hence, the principal value of cosec-1 (-1) is equal to
−π

46. Principal value branch of sec1 x is [0, π

2
)∪ (
π

2
, π] and its graph is shown below.

47. Let tan-1(-1) = y

⇒ tan y = - 1
π
⇒ tan y = − tan
4

my
π
⇒ tan y = tan(− )
4

Since, the principal value branch of tan-1 is [− π

2
,
π

2
] .
Therefore, principal value of tan-1(-1) is − . π

48. We know that the range of the principal-value branch of tan-1 is (

de
−π π
, )
2 2

tan-1 (√3)= θ
–
Let , Then, we have,
– π π −π π
tan θ = √3 = tan ⇒ θ = ∈ ( , )

Hence, the principal value of tan-1

3 3 2

–
(√3)
2

is equal to
a π
ac
3

49. Let cos −1

(
−1

2
) = y

1
⇒ cos y = −
2
π
⇒ cos y = − cos
th

π 2π
⇒ cos y = cos(π − ) = cos
3 3

Since, the principal value branch of cos-1 is [0, π] .

rar

−1
Therefore, principal value of cos −1
(
2
) is 2π

3
.

50. We know that, for any x ∈ R, cot-1x represents an angle in (0, π )

– –
(− √3) = An angle in (0, π ) whose cotangent is (− √3)
ast

−1
cot
π
= π −
6
5π
=
6
–
−1 5π
.
Sh

∴ cot (− √3) =
6

51. Principal value branch of cot-1 x is (0, π ) and its graph is shown below.

52. Principal value branch of sin1 x is [− π

2
,
π

2
]

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Er.anurag sir's classes
 And its graph is given here

53. we know that Principal value branch of cosec-1 x is [− π

2
, 0) ∪ (0,
π

2
] and its graph is given below.

my
de
54. Given sin −1
(
1

3
) − cos
−1
(−
1

3
)

We know that cos-1 (-θ) = π - cos-1 θ a

−1 1 1
ac
−1
= sin ( ) − [π − cos ( )]
3 3

−1 1 −1 1
= sin ( ) − π + cos ( )
3 3

−1 1 −1 1
= sin ( ) + cos ( ) − π
3 3
th

π
= − π
2
π
= −
2
rar

Therefore we have,
−1 1 −1 1 π
sin ( ) − cos (− ) = −
3 3 2

√3
55. Let cos −1
(
2
) = y
ast

√3
⇒ cos y =
2
π
⇒ cos y = cos
6
π
⇒ y =
Sh

Since, the principal value branch of cos-1 is [0, π] .

√3
Therefore, principal value of cos −1
(
2
) is π

6
.

56. Let us consider tan-1(1) = x then we obtain

π
tan x = 1 = tan 4

We know that range of the principle value branch of tan-1 is [− π

2
,
π

2
]

Thus, tan-1 (1) = π

Let cos −1
(−
1

2
)= y

1 π 2π
cos y = − 2
= cos(π −
3
) = cos(
3
)

We know that range of the principle value branch of cos-1 is [0, π ]

1 2π
Thus, cos −1
(−
2
)=
3

sin-1 (− ) =
1
Let 2
z

π π
sin z = − sin = sin(− )
6 6

We know that range of the principle value branch of sin-1 is [− π

2
,
π

2
]

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Er.anurag sir's classes
 Thus, sin-1 (− 1

2
)= −
π

Now,we have
−1 −1 1 1
tan (1) + cos (− ) + sin(− )
2 2
3π+8π−2π
= π

4
+
2π

3
−
π

6
=
12
=
9π

12
=
3π

57. (d) tan-1( 1

2
)

Explanation: tan-1( 1

2
)

58. (c) tan-1 ( 4

3
)

Explanation: tan-1 ( 4

3
)

(b) tan-1 (
11
59. 2
)

Explanation: tan-1 ( 11

2
)

60. (b) tan-1 ( 1

8
)

Explanation: tan-1 ( 1
)

my
8

61. (b) R, (− π

2
,
π

2
)

π π
Explanation: R, (− 2
,
2
)

a de
ac
th
rar
ast
Sh

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Er.anurag sir's classes