Class: XII
Subject:
√ Maths
Topic: Inverse Trigonometric Functions
No. of Questions: 25
–1
Q1. If sin x =
,x
–1
(–1, 1), then cos x =
ns
A.
B.
D.
–1
–1
–1
−1
,
]
B. [−1, 1]
C. (2, 2ᴨ]
D. [−
,
x is
[Foreign 2010]
]
as
A. [−
–1
- sin x
kI
Q2. The domain of sin
Or cos x =
IT
Ans. A sin x + cos x =
ia
C. –
Right Answer Explanation: B
Fact
Page|1
−1
Q3.tan (cos x) is equal to
A.
B.
C.
D.
−1
D.
Ans. B
tan (cos x) = tan
=
[Hots; Delhi 2011, 2009; All India 2009]
kI
as
C.
-1
=
is equal to
A.
B.
tan
ia
Q4. cos
x = cos
IT
-1
Let cos x =
ns
Ans. A
Range is [0, ᴨ]. So first convert 7 ᴨ/6 into that range.
Page|2
Q5. If sec
-1
, then x + y + z is equal to
A. xyz
B. 2xyz
2
C. xyz
2
D. x yz
Ans. A
-1
-1
-1
ns
-1
tan x + tan y + tan z =
-1
-1
2
Q6. 2 cos x = cos (2x – 1) holds true if
IT
A. |x| ≤ 1
x + y + z = xyz
ia
sec
C. |x| <
D. None of these
kI
B. 0 ≤ x ≤ 1
as
[Ans. B The result holds true only if 0 ≤ 2 cos-1x ≤ π i.e. if 0 ≤ cos-1 x ≤
, i.e. if 1 ≥ x ≥ 0
Q7. tan (sin1 x) is equal to
A.
B.
C.
D. None of these
Page|3
Ans. A tan (sin1x) =
Q8. cos1
is equal to
A. 
B.
ns
C.
= cos1
IT
Ans. B cos1
ia
D. None of these
kI
Q9. If sin1 x = /6, then cos1x is equal to
A.
C.
as
B.
D. None of these
Ans. B cos1 x = −
=
Page|4
Q10. If θ = tan1 x, then sin 2θ is equal to
A.
B.
C.
D. None of these
ns
Ans. A Given θ = tan1 x, Therefore, sin 2θ =
ia
Q11. The range of tan1 x is
IT
A. (, -π)
B.
kI
C. (, π)
D.
as
Ans. B Range of tan1 x is
Q12. cot (cos1x) is equal to
A.
B.
C.
D. None of these
Page|5
Ans. B cot (cos1x) =
=
Q13. cos (tan1x) is equal to
A.
B. 
C.
A. 1
ia
=
is equal to
as
B. tan 1
C.
∴ cos (tan1 x) = cos θ =
kI
Q14. If x > 0, then tan1 x + tan1
<θ<
IT
Ans. A Let tan1x = θ ⇒ x = tan θ, 
ns
D. None of these
D. None of these
Right Answer Explanation: C
tan1 x + tan1
= tan1 x + cot1 x =
Page|6
Q15. If x ∈ [1, 1] and sin1 x =
, then cos1 x is equal to
A.
B.
C.
D. None of these
sin1 x =
−1
1
(1 + x) + tan (1 − x) =
is
ia
Q16. A solution of the equation tan

ns
Ans. D cos1 x =
IT
A. x = 1
B. x = − 1
kI
C. x = 0
D. x =
as
Ans. C
-1
-1
tan (1 + x) + tan (1 - x) =
-1
tan
-1
tan
=
= tan
2
x =0
=
x=0
Page|7
Q17. The value of sin
−1
is
A. 0
B.
C.
D.
ns
Ans. A
−1
= sin
-1
= sin
-1
=0
IT
-1
kI
= sin
ia
sin
So, first option is the answer.
as
Q18.Directions: The following question has four choices, out of which ONLY ONE is correct.
A function f: R
R describes a curve y = f(x). A point P(x, y) lies on the curve and satisfies
the equation
A.
0
B.
1
C.
2
D.
None of these
The number of asymptote(s) of the curve is (are)
Page|8
Ans. C
The given equation is
(x - 3) (y + 3) = - 10
By graph
IT
ia
ns
Graph of (x - 3) (y + 3) = - 10
The number of asymptotes = 2
as
kI
Q19. Directions: The answer to the following question is a single digit integer, ranging from 0 to
9. A polygon is obtained by joining the points (taken in order) on the curve
and its area is recorded. Now another polygon is obtained by joining the
midpoints of the sides of the above polygon taken in the same order and its area is also
recorded. The process continued indefinitely. If A is the sum of all such areas, the value of
, where [.] denotes g.i.f., is
Right Answer Explanation: 6
is well defined, if
Page|9
The points are
If
is the area of the polygon A1A2A3A4, then
If
is the area of the polygon obtained by joining the mid points of the sides of the above
polygon, then
=2
.
= .
Q20. Write the principal value of
cos
+ cos
√
+
− cos
[∴ cos (− ) =
= + − =
[Delhi 2013C]
]
−√3 .
(1)
[All India 2013]
as
Q21. tan-1 √3 − cot
− cos
=
IT
= cos
√
(− ) .
kI
cos
+ cos
ia
Ans.
√
ns
Hence, some of all such areas A =
Ans.
tan-1 √3 − cot (− 3)
= tan ( 3) − { −
( 3)}
[∴ Principal value of cot-1 is] 0, π [∴ cot-1(–x) = π–cot-1 x]
= tan ( 3) − + cot √3
= −
=−
∴ tan
+ cot
=
Page|10
Ans.
cos
− 2 sin
= cos
[∴
−
cos
sin-1
= −2 −
[∴ cos-1 (
= + =
−
cos
− ,
)=
Q23. tan-1√3 − sec
[Delhi 2012]
−
− 2 sin
And that of
.
(0, )]
sin (
)= ]
(−2).
tan
∴ tan = √3
= −
=
[∴ tan (
Q24.
sin-1
− sec
sec
2
sec = −2
as
= tan
(12)
kI
3 − sec
[All India 2012]
IT
Ans.
We know that, the principal value for tan-1 x is
− , and that of sec-1 x is [0, π]−
.
So, tan
ns
− 2 sin
ia
Q22. cos-1
)=
sec
(
)= ]
-√
Ans.
We know that, the principal value of sin-1
− , .
∴ sin-1 −
√
= sin
− sin
is
Page|11
=− ∈ − ,
∴ sin-1 −
√
−
=−
∴
[∴
Q25. If tan-1 √3 + cot -1x=
−
sin-1
then find the value of x.
√3 = tan
+ cot
=
(1)
[All India 2010C]
as
kI
IT
On equating, we get
= √3
∴ tan
(1)
ia
⟹ tan
√3 = − cot
sin(− ) = −
(sin ) = ]
Ans.
Given that, tan-1 √3 + cot -1x=
⟹ tan
= − sin
ns
= sin
Page|12