Inverse Trigonomegftric Functions Question (1).pptx

Mathematics
Lecture -
01
By – Ashish Agarwal
Sir

1. Graphs of y = f(x) and y = f–1(x) are mirror images of each other about line y = x.
(T/F)
2. If ( ,
α )
β lies on grape of y = f(x) then ( ,
β )
α lies on the graph of y = f–1(x).
(T/F)
3. All solutions of f(x) = f–1(x) lie on the line y = x. (T/F)
4. All solutions of f(x) = f–1(x) lie on the line y = x or on a line with slope –1.
(T/F)
5. Solutions of f(x) = f–1(x) not lying on the line y = x can only arise in cases where (a,
b) & (b, a) a ≠ b both lie on graph of y = f(x). (T/F)

6. Solutions of f(x) = f–1(x) not lying on the line y = x can only arise in cases where
f(a) = b & f(b) = a, a ≠ b, a, b ∈ Df. (T/F)
7. Solutions of f(x) = f–1(x) not lying on line y = x can only arise in case graph of y = f(x)
crosses y = x. (T/F)
8. Solutions of f(x) = f–1(x) not lying on the line y = x will always arise in cases where (a,
b) & (b, a), a ≠ b both lie on grape of y = f(x). (T/F)
9. For a function to be invertible it should be

= x ∀ x ∈
A
10. If f: A → B is a bijection and g: B → A be a function such that g
f x
then g =
= x ∀ x ∈
B
11. If f: A → B is a bijection and g: B → A be a function such that f
g x
then g =

Write explicitly, functions of y defined by the following equations and also find the
domains of definition of the given implicit functions :
(a) 10x + 10y = 10 (b) x + |y| = 2y
QUESTION

= x 4 − x then find f−1(x) and solve the equation,
f: 2, ∞ → −∞, 4 ,where f
x f−1(x) = f(x).
QUESTION

f: 1
, ∞ → 3
,
∞
2 4
is the function given by f(x) = x2 − x + 1. Show that f is bijective,
Find
2 1
2
f−1. Hence solve for real x, x − x + 1 =
+
x
−
3
4
QUESTION

x +
p2,
Let a function f defined from R → R as f(x) = ቈ
px + 5,
for x ≤
2
for x >
2
If the function is surjective, then find the sum of all possible integral values of p in [–
100, 100].
QUESTIO
N
MB 2

Find the following:
2
(i) sin π
−
sin−1
−
32
(ii) cos
cos−1
2
− 3
+
π
6
(iii) tan−1 tan
3π
4
QUESTION

cos−1 cos 2
cot−1
2 − 1 is equal to
π
4
3
π
4
none of these
A 2 −
1
B
C
D
QUESTIO
N

The value of sin−1 cot
sin−1
2
− 4
3
+
cos−1
4
12
+ sec−1 2 is
0
π
2
π
3
none of these
A
B
C
D
QUESTIO
N

A lion moves in the region given by the graph y − |y| − x + |x| = 0. Then on which of the
following curve a person can move so that he does not encounter lion -
y = e−
x
1
y =
x
y = signum
(x)
y = − 4 +
x
A
B
C
D
QUESTIO
N

QUESTION
1. Fill in the Table
[Ans. D]
f(x) g(x) f(x) + g(x) f(x) − g(x) f(x) ⋅ g(x)
f x
g x g ∘ f x (f ∘ g)(x)
odd odd
even even
odd even
even odd
2. Suppose that f is an even, periodic function with period 2, and that f(x) = x for all x
in interval [0, 1]. Find the value of f(3.14).
x+
1 n
times
3. Let f(x) = x−3
, x ≠ −1. Then f 2010(2014) [where f n(x) = f ∘ f
… .∘ f (x)] is:
(A) 2010 (B) 4020 (C) 4028 (D) 2014
[Ans. 0.86]
KTK

4. Find the inverse of f(x) = 2log10 x + 8 and hence solve the equation f(x) =
f−1(x).
5. For a > 0, if f(x + a)
=
1
2
2
+ f(x) − f(x), prove that f is periodic.
QUESTIO
N
KTK

1. Find the principal value of each of the following:
1
3
(i) tan−1 (ii)
tan−1
−
1
3
(iii) tan−1 cos
π
2
(iv) tan−1 2 cos
2π
3
2. For the principal values, evaluate each of the following:
(i) tan−1 −1 +
cos−1
−
1
2
(ii) tan−1 2 sin 4
cos−1
3
2
3. Evaluate each of the following:
(i) tan−1 1 +
cos−1
−
1
2
+
sin−1
−
1
2
(ii)
tan−1
−
1
3 −
π
2
(iii) tan−1 tan
5π
6
+ tan−1 − 3 + tan−1
sin
+ cos−1 cos
13
π6
Bumper Practice Problems
(BPP)

π
6
(ii) −
π
6
(iii) 0 (iv) −
π
4
π
2
(ii)
π
3
1. (i)
2. (i)
3. (i)
3
π
4
(ii) −
3π
4
(iii) 0
BPP
Answers

Mathematics
Lecture -
02
Sir

1. ITF Domain Range
sin−1 x
cos−1 x
tan−1x
cot−1 x
sec−1 x
cosec−1x
2. sin−1 x |max = ,
sin−1 x |min =
cos−1 x |max = ,
cos−1 x |min = ,

3. ITF Inc. Dec. Even Odd
𝐬in−1x
cos−1 x
tan−1 x
cot−1 x

4. sin−1 −x =
tan−1 −x =
5. Domain of f x = sin−1 x + tan−1 x +
sec−1 x is
6. cos–1 (0.3) > cos–1 (0.5) (T/F)
sin–1 (0.3) > sin–1 (0.2)
cot–1 (5) > cot–1 (2)
tan–1 (7) < tan–1 (3)
(T/F)
(T/F)
(T/F)

7. sin−1 1 = , cos−1 1 = , cot−
1
3 = ,
tan−1 3
=
,
sec−1
2
3
=
8. If f is increasing bijective function then solutions of f(x) = f−1(x) can only lie on
line y = x. (T/F)
9. lim tan x =
x→∞
lim cot x =
x→∞
lim tan−1 x =
,
x→−∞
lim cot−1 x =
x→−∞

Nichod !!
S. No. f(x) Domain Range
1 sin−1x |x| ≤ 1
π π
− ,
2 2
2 cos−1x |x| ≤ 1 [0, π]
3 tan−1x x ∈ R
π π
− ,
2 2
4 sec−1x |x| ≥ 1 [0, π] − π
or 0, π
∪ π
, π
2 2
2
5 cosec−1x |x| ≥ 1
π π
− , − {0}
2 2
6 cot−1x x ∈ R (0, π)

Find the following:
2
(i) tan cos−1 1
+
tan−1
−
13
(ii) cos−1 cos
7π
6
(iii) cos tan−1
3
4
(iv) tan sin−1 3
+ cot−1
3
5
2
QUESTION

9
16
25
23
The value of tan2 sec−1
3
+
cot2
cosec−1 4 is -
QUESTIO
N
A
B
C
D

Evaluate:
1. sin arc
sin
3
5
− arc
cos
3
5
2. tan 1
cos−1
5
2
3
3. sin tan−1 cos
cot−1
−
1
3
4. sin tan−1 2 + cos
tan−1 2
5. tan 2 tan−1 1
− π
5
4
QUESTIO
N

π
2
π
3
π
6
π
4
tan−
1
1 +
3
3 +
3
+
sec−1
8 + 4
3
6 + 3
3
is equal to :
QUESTION [JEE Mains 2023
(Jan)]
A
B
C
D

(a)
sin−1
−
1
2
+
cos−1
−
1
2
−
tan−1(−
3) +
cot−1
−
1
3
(b) sin 2 sin−1
3
(c)
5
cos 2 tan−1
2
+ sin 2 tan−1
3
QUESTION

1
7
2 2 −
1
7 −
1
1
2
2
4
A possible value of tan 1
sin−1
6
3
8
is :
(Feb)]
A
B
C
D

(a) Domain & range of cos−1
x
ex
(b)
(c)
(d)
Domain & range of y = sin−1
Domain & range of cos−1 x
Domain & range of cot−1
sgnx
(e) Domain & range of
1
(f) Domain & range of
ln cot x
1
sgn sec−1
ex
QUESTIO
N

The domain of definition of the function f
x
= sin−1
log
3
x
3
is
[1, 9]
[–1, 9]
[–9, 1]
[–9, –1]
A
B
C
D
Ans. A
QUESTIO
N

[0,
2π]
[0,
2π)
[0,
π)
[0,
π]
The range of f x = 4
sin−1
x2
x2 +
1
is
QUESTION [JEE Mains 2023 (13
April)]
A
B
C
D

3x2 + bx + 3 , x ∈ R is
0,
π
2
, then square of sum of
If range of the function f(x) =
tan−1
all possible values of b will be
0
18
72
None of these
A
B
C
D
QUESTION

Solve:
(1) log2 tan−1 x >
1
(2) cot−1 x 2 − 5 cot−1 x + 6
> 0
QUESTIO
N

The number of real roots of the equation
4
tan−1 x(x + 1) + sin−1x2 + x + 1 = π
is :
1
2
4
0
(July)]
A
B
C
D

Let f(x) = a + b cos−1 x (b > 0). If domain and range of f(x) are the same set then (b
− a)
is equal to
QUESTIO
N
A 1 −
1
π
2
+ 1
π
2
1 −
π
2
B
C
D

(−∞,
0)
(−∞,
0]
(0,
∞)
[0,
∞)
f ∶ R → R is defined as f(x) = ቈx2 + 2mx
− 1
for x ≤
0.
mx − 1
for x > 0
If f(x) is one-one then m must lies in the interval
Mind
Bender
A
B
C
D

1. The number of solutions of equation 2 sin−1 x 2 − 5 sin−1 x + 2 = 0 is
(A) 2 (B) 1 (C) 3 (D) None of
these
2. sec−1 sec
−30∘
(A) –60°
is equal to
(B) –30° (C) 30° (D) 150°
3. sin 2 sin−1
0.8
(A) 0.96
is equal to
(B) 0.48 (C) 0.64 (D) None of these
QUESTIO
N
[Ans. A]
[Ans. B]
[Ans. C]
KTK

1. Evaluate each of the following:
(i) sin sin−1
7
2
5
(ii) sin cos−1 5
1
3
(iii) sin tan−1
24
7
(iv) cosec cos−1
3
5
(v) sec sin−1
12
1
3
1
7
(vii) cot cos−1
3
5
(vi) tan cos−1 8
(viii) cos tan−1
24
7
2. Prove the following results:
(i) tan cos−1 4
+ tan−1
2
5
3
=
17
6
(ii) cos sin−1 3
+ cot−1 3
=
5
2
6
5
13
(iii) tan sin−1 5
+ cos−1
3
13
=
63
1
6
(iv)
sin
cos−1 3
+ sin−1
5
5
=
63
6
5
Basic Practice Problems
(BPP)
[CBSE 2012]

1. (i)
7
(ii)
1
2 (iii)
2
4 (iv)
(v)
2
5
1
3
5
(vi)
1
3
1
5
8
(vii)
2
5
3
4
(viii)
5
4
7
2
5
BPP
Answers

Mathematics
Lecture -
03
Sir

1. tan−1 x > 1 ⇒ ⇒ x ∈
3
4
2. sin−1 x > ⇒ ⇒ x ∈
1
4
3. cos−1 x > ⇒ ⇒ x ∈
3
2
4. cot−1 x > ⇒ ⇒ x ∈
5. 2 < cot−1 x < 3 ⇒ x ∈
6. −1 < tan−1 x < 1 ⇒ x ∈

7. cot−1 x 2 − 4 cot−1 x − 5 > 0 then x ∈
8. 6 sin−1 x 2 − 5 sin−1 x + 1 < 0 then x ∈
9. cos−1 x 2 − 5 cos−1 x + 6 > 0 then x ∈
10. Range of f
x
= cos−1x + cot−1 x + cosec−1x is

1
2
11. sin−1 x = ⇒ x = 2
sin−1 1
=
tan−1 x = π
⇒ x
=
2 + 1
=
8
cos−1 x = 2 ⇒ x
=
1
2
=
tan−1
cos−1
cot−1 3 =
cot−1 x = 4 ⇒ x =
3
4
sin−1 x = − ⇒ x = co
s
−
1
1
2
=
tan−1 1 =
cot−1 x = 3 ⇒ x =
tan−1 x = 1 ⇒ x =
cos−
1
−
1
2
=
cos−1 x
=
1
2
⇒ x
=
cot−1 x = 3 ⇒ x =
1
2
cos−1 x = − ⇒ x =

Range of the function f(x) =
cos−1
1
ex+e−x
is
0,
π
,
,
π
π
6
2
π
π
3
2
,
π
2π
2
3
A
B
C
D
QUESTIO
N

The domain of the function cosec−1
1 +
x x
is :
−1, −
1
2
∪ (0,
∞)
1
− , 0 ∪ [1,
∞)
2
1
− , ∞ −
{0}
2
1
− , ∞ −
{0}
2
QUESTIO
N
A
B
C
D

Let x denote the greatest integer less than or equal to x. Then the domain of
f x = sec−1 2 x + 1 is:
−∞, ∞
−∞, ∞ − 0
−∞, −1 ∪ 0,
∞
−∞, −1 ∪ 1,
∞
A
B
C
D
Ans. A
Jan)]

If cos−1 x + cos−1 y + cos−1 z = 3π then the
value of
x2014 + y2014 + z2014
+
6
x2013 + y2013 +
z2013
QUESTIO
N

Find the values of:
(i) sin−1
sin 3
(ii) sin−1
sin 53
(iii) sin−1
sin 7
(iv) sin−1
sin 10
(v) sin−1
sin 20
QUESTIO
N

Find the values of:
(i) cos−1
cos 2
(ii) cos−1
cos 3
(iii) cos−1
cos 5
(iv) cos−1
cos 7
(v) cos−1
cos 10
QUESTIO
N

Find the values of:
(i) tan−1
tan 3
(ii) tan−1
tan 5
(iii) tan−1
tan 7
(iv) tan−1
tan 10
(v) tan−1
tan 15
QUESTIO
N

Mathematics
Lecture -
04
Sir

1. Function Period Yaad Rakho
sin−1(sin x) sin−1(sin x) = ൝
cos−1(cos x) cos−1(cos x) = ൝
tan−1(tan x) tan−1(tan x) = ൝

2. (a) y = sin–1 x (b) y = cos–1 x
(c) y = tan–1 x (d) y = cot–1 x

3.
sin−1
1 1
2
2
= = cos−1 hence graph of y = sin–1 x and y = cos–1 x
intersect at x = , y =
4.
sin–1 x > cos–1 x has solution
sin–1 x < cos–1 x has solution

5. tan−1 1 = = cot−1 1 hence graph of y = tan–1 x and y = cot–1 x
intersect at x = , y =
6.
(a) tan–1 x > cot–1 x if x 
(b) tan–1 x < cot–1 x if x 

cos–1 (cos 100) =
7. sin–1 (sin 100) = tan–1 (tan
100) =
8. (a) y = sin–1 (sin x)

8. (b) y = cos–1 (cos x)
(c) y = tan–1 (tan x)

9. sin−1 x
=
1
2
⇒ x
∈
1 < tan−1 x < 2 ⇒ x ∈
cos−1 x > 2 ⇒ x ∈
tan−1 x > 2 ⇒ x ∈
cot−1 x > 2 ⇒ x ∈

Integral solution of the inequality 3x2 + 8x < 2
sin−1
sin 4 −
cos−1
cos 4 is
1
–1
0
2
A
B
C
D
QUESTIO
N

If x = sin–1 (sin10) and y = cos–1 (cos10), they y – x is equal to
QUESTION [JEE Mains
2019]
A. 10
B. 0
C. 
D. 7


If a = sin−1sin 5 and b =
cos−1
cos 5 , then a2 + b2 is equal
to
Jan)]
Ans. C
25
4π2 +
25
8π2 − 40π +
50
4π2 − 20π +
50
A
B
C
D

3π −
11
4π −
9
4π −
11
3π +
1
cos−1 cos −5 + sin−1 sin 6 − tan−1 tan 12 is equal to:
(The inverse trigonometric function take the principals values)
QUESTION [JEE Mains
2021]
A
B
C
D

Value of sin−1 sin 7 +
cos−1
cos 13is
21 −
5π
20 −
6π
20 −
5π
21 −
6π
A
B
C
D
QUESTIO
N

10
π7
+ cot−1
cot
26
π5
(a) Solve : cos−1
cos
(b) Prove that cos−1
cos
1000π
7
+
sin−1
sin
601π
1
4
=
13π
1
4
QUESTION

Show that:
sin−
1
sin
33π
7
+
cos−1
cos
46π
7 8
+ tan−1 − tan 13π
+ cot−1
cot
−
19π
8 =
13π
7
QUESTION

α
α −
2
α +
2
2 −
α
If f(x) = x11 + x9 − x7 + x3 + 1 and f sin−1(sin 8) =
α, α is constant, then
f tan−1(tan 8) is
equal to
QUESTIO
N
A
B
C
D

Let f: [0, 4π] → [0, π] be defined by f(x) = cos−1(cos x). The number of points x ∈
[0, 4π]
1
0
satisfying the equation f(x) = 10−x
is
QUESTION [JEE Advanced
2014]

Let f x = x + 1 x + 2 x + 3 x + 4 + 5 where x ∈−6, 6 . If the range of the
function is a, b where a, b ∈ N then find the value ofa + b .
Ans. 5049
Mind Bender

1. If cos−1 x + cos−1 y + cos−1 z = 3π, then find the
value of
x2013+y2013+z2013+
6
x2014+y2014+z201
4
sin−1 w + sin−1
z
= π2 then compute the value of
2. If sin−1 x +
sin−1 y (x + y)(w
+ z).
QUESTION
[Ans. 1]
[Ans. 4]
3. Let 'f' be an even periodic function with period '4' such that f x = 2x − 1, 0 ≤ x
≤ 2.
The number of solutions of the equation f(x) = 1 in [–10, 20] are
[Ans. 15]
KTK

4. If x1, x2 and x3 are the positive roots of the equation x3 − 6x2 + 3px − 2p = 0, p
∈ Rthen the value of sin−1
x1 x2 x2
x3
1
+
1 1
+
1 1
+
1
x3
x1
+ cos−1 − tan−1 is equal to
(A) π
4
(B) π
2
(C) 3π
4
(D)
π
5. The domain of the function f(x) =
1
logπ sin−1 x
−1
4
, is
(A) −1, 1
(B) 0,
1
2
2
(C) 0,
1
2
(D) −1,
1
2
x2 − 5x −
11
2
= π, then
6. If α and β are the two zeroes of the equation 3
cos−1
α3 + β3 equals
(A) 255 (B) 215 (C) –215
(D) –217 [Ans. B]
QUESTIO
N
KTK
[Ans. A]
[Ans. B]

[Ans. 2π −
4]
1. sin−1 sin 1 + sin−1 sin 2 + sin−1 sin
3
2. sin−1 sin 10+ sin−1 sin 20+ sin−1
sin 30
+ sin−1 sin
40
[Ans. 0]
[Ans. 2π +
2]
[Ans. 20 −
4π]
[Ans. 7π −
20]
[Ans. 0]
3. cos−1 cos 1 + cos−1 cos 2 + cos−1 cos 3 + cos−1 cos
4
4. cos−1 cos 10+ cos−1 cos 20+ cos−1 cos 30+ cos−1
cos 40
5. sin−1sin 10 + cos−1 cos 10
6. sin−1sin 50 + cos−1 cos 50
7. sin−1sin 100 + cos−1 cos 100
[Ans. 0]
Bumper Practice Problems
(BPP)

Mathematics
Lecture -
05
Sir

1. cot–1 x = ቐ
if x >
0
,
if x <
0
tan–1 x = ቐ
if x >
0
ifx <
0
2. sec–1 x = , if |x|
cosec–1 x = , if |x|
3. sin–1 x = , if |x|
–1

4. cos–1 (–x) = , sec–1 (–x) =
cot–1 (–x) =
5. sin–1 (–x) = , tan–1 (–x) =
cosec–1 (–x) =
6. sin–1 x + cos–1 x = , tan–1 x + cot–1 x =
sec–1 x + cosec–1 x =

7. y = sec–1 x y = cosec–1 x

8. y = sec–1 (sec x) Domain =
Range =
9. y = cosec–1 (cosec x)
Domain =

10. y = cot–1 (cot x) 11. y = tan–1 (tan x)

Find x
if
1. 4 sin−1 x + cos−1 x =
3π
4
2. 5 tan−1 x + 3 cot−1 x
= 7π
4
QUESTIO
N

(a) Solve for x : 2 sin–1 x + 3 cos–1 x =
3π
2
(b) Solve for x : cot–1 x + 2 tan–1 x =
3π
4
QUESTIO
N

Find the range of
8
1. f(x) = sin−1 x + cos−1 x + tan−1 x
2. f(x) = sin−1 x + tan−1 x + sec−1 x
2
3. If tan−1 x 2 + cot−1 x2 = 5π
then find the value
of x.
QUESTIO
N

Maximum & Minimum value of
f x = sin−1 x 3 +cos−1 x3
QUESTION

Using the principal values of the inverse trigonometric functions, the sum of the
maximum and the minimum values of 16 sec−1 x 2 +cosec−1x 2is:
24π
2
18π
2
22π
2
31π
2
A
B
C
D
Ans. C
Jan)]

The set of all values of k for which tan−1 x 3 +cot−1 x 3 = kπ3, x ∈ R, is the
interval :
1
,
7
32
8
1
,
13
24
16
1
,
13
48
16
1
,
9
32
8
(June)]
A
B
C
D

4 1 − x2 1 − 2x2
4x 1 − x2 1 −
2x2
2x 1 − x2 1 −
4x2
4 1 − x2 1 − 4x2
If 0 < x
<
1
2
−1 −1
and sin x
= cos x
, then the value of sin
𝝰 β
2π
𝝰
𝝰+
β
is
QUESTION [JEE Mains
2022]
A
B
C
D

−1 −1 −1
If sin x
= cos x
= tan y
; 0 < x < 1 then the value of
cos
a b c
π
c
a+
b
is
1 − y2
y y
1 − y2
1 −
y2
1 +
y2
1 −
y2
2
𝑦
QUESTION [JEE Mains
2021]
A
B
C
D

tan−1
α
+
cosec2
cot−1
β
= 36, then α2 +
β
If for some α, β; α ≤ β, α + β = 8 and
sec2
is
Ans. 14
Jan)]

The value of cot−1
1 + tan22 −
1 tan
2
−
cot−1
1 +
tan2
1
2
+
1
ta
n
1
2
is equal to
3
π −
2
5
π +
2
5
π −
4
3
π +
2
A
B
C
D
Ans. C
April)]

Prove that the function defined as, f(x) = ൥e−
ln
x
ln
x
− {x}
;
x
;
where ever it exists
otherwise, then
f(x) is odd as well as even. (where {x} denotes the fractional part function)
Mind
Bender

π
,
π
6
4
defined by f(x) =
tan−1
x2+1
x2+
3
. Then f(x)
is
(A) injective and surjective
(B) injective but not surjective
(C) surjective but not injective
(D) neither injective nor surjective
2. The value of 3 sin 1
arc cos 1
+ 4 cos
1
arc cos 1
2 9 2
8
is equal to
(A) 5 (B) 4 (C) 1 (D) 0
1
1+cos2 x
3
= pπ
have
a
3. The true set of values of p for which the equation cos−1
solution is
(A) [0, 1] (B) [0, 2] (C) [1, 2] (D) [1, 3/2]
QUESTION
1. Let f ∶ R
→
[Ans. C]
[Ans. A]
[Ans. A]
KTK

4. The value of p ∈ R for which the equation
sin−1 log10 x 2 − 2 log10 x + 2 + tan−1
log10 x 2 − 2 log10 x + 2 + cos−1
log10 x 2 − 2 log10 x = p possess solution is
(A) 5π
4
(B) 3π
(C) 3π
(D) 7π
4 2
4
6. Consider a real-valued function f
x
The range of f(x) is
(A) 0, 3 (B) 1,
3
= sin−1 x + 2 +1 − sin−1
x.
(C) 1, 6
(D) 3,
6
QUESTIO
N
[Ans. D]
5. Consider a real-valued function f x = sin−1 x + 2 +1 − sin−1 x.
The domain of definition of f(x) is
(A) [–1, 1] (B) [sin 1, 1] (C) [–1, sin 1] (D) [–1, 0]
[Ans. C]
KTK
[Ans. D]

Mathematics
Lecture -
06
Sir

1. sin–1 x + cos–1 x =
2. tan–1 x + cot–1 x =
3. sec–1 x + cosec–1 x =
4. cot–1 x =
if x > o cot–1 x =
if x < o
5. tan–1 x =

6. tan–1 x + tan–1 y = if
= if
7. sin–1 (sin x) = if π
< x ≤
3π
2
2
8. tan–1 (tan x) = if π
< x <
3π
2
2
9. cos–1 (cos x) = if −π ≤ x <
0

10. tan–1 x – tan–1 y = if x, y > 0.
11. f(x) = sec–1 x + cosec–1 x + tan–1 x then range of f(x) is
12. f(x) = sin–1 x + cos–1 x + cot–1 x then range of f(x) is
x
13. Graph of y = x +
1
is
14. y = sin–1 (sin (–5)) + cos–1 (cos (–10)) then value of y is

If the inverse trigonometric functions take principal values then
1
0
cos−1 3
cos
tan−
1
4
3
+
2
5
sin
tan
−
1
4
3
is equal to:
June)]
Ans. C
0
π
4
π
3
π
6
A
B
C
D

50 tan 3
tan−1
1
1
2
5
+ 2 cos−1 +
4
2
2 tan 1
tan−1
2 2 is equal to
June)]
Ans. 29

Considering only the principal values of the inverse trigonometric functions, the value
of tan
sin−1
3
2
5
5
− 2 cos−1 is
QUESTION [JEE Advanced 2024 (Paper
2)]
Ans. B
7
24
−
7
2
4
−
5
2
4
5
2
4
A
B
C
D

cosec 2 cot−15 +
cos−1
4
5
is equal to:
Feb)]
Ans. B
7
5
5
6
6
5
5
6
5
6
3
3
6
5
3
3
A
B
C
D

5
tan 2 tan−1 1
+
sec−1
5
+ 2 tan−1
1
2 8
is equal to :
July)]
Ans. B
2
1
4
5
4
A 1
B
C
D

15x2 − 8x − 7
= 0
5x2 − 12x + 7
= 0
25x2 − 18x − 7
= 0
25x2 − 32x + 7
= 0
Let α = tan 5π
sin
2cos−1
1
16
5
and β = cos
sin−1
4
5
+
sec−1
5
3
where the inverse
trigonometric functions take principal values. Then, the equation whose roots are α and β
is :
(June)]
A
B
C
D

Find the solution of following equations :
5
(i) 2 cot−1 2 − cos−1 4
=
cosec−1 x
QUESTION

Find the solution of following equations :
(ii) cos−1 x − sin−1 x = cos−1(x
3)
QUESTION

cos sin−1 3
+ sin−1 5
+
sin−1 33
5 13
65
is equal to:
3
3
6
5
1
3
2
6
5
A
B
C
D 0
Ans. D
Jan)]

The domain of the function f(x) =
sin−1
|x|+5
x2+1
is (−∞, −a] ∪ [a, ∞). Then a is equal
to:
2
17
+
1
17−
12
1
7
2
1+
172
QUESTION [JEE Mains 2020]
A
B
C
D

1. If the equation 5 arc tan x2 + x
+ k
solutions, then the range of k, is
(A) 0, 5
(B) −∞,
5
4
4
(C) 5
, ∞ (D) −∞,
5
4
4
2. If
sin−1
3
9
4 6
x2 − x
+ x
… +
cos−1
3, then
number of values of 'x' is equal to
8 12
x4 − x
+ x
… = π
, where 0 ≤ |x|
<
3 9 2
(C) 3 (D) 4
(A) 1 (B) 2
3. A value of α for which sin
cot−1
1 + α = costan−1 α ,
is
(A) −1
2
(B) 0 (C) 1
2
(D) 1
b
2
cos−1 x 2 is equal to aπ
(a and b are coprime),
4. If maximum value of sin−1 x 2
+
then (a + b) equals (C) 4 (D) 9
QUESTIO
N
[Ans. B]
[Ans. C]
KTK
+ 3 arc cot x2 + x + k =
2π, has two distinct
[Ans. A]
[Ans. D]

5. Column I contains functions and column II contains their range. Match the entries of
column I with the entries of column II.
Column-I Column-II
x
(P) 0,
π
1+
x
x (Q) π
,
3π
4
4
1+
x
x
(R) −
,
π
π
4
4
(A) f x =
sin−1
(B) g x =
cos−1
(C) h x =
tan−1
(D) k x =
cot−1
1+
x
x
1+
x
(S) −
,
π
π
2
2
QUESTIO
N
[Ans.(A) S, (B) P, (C) R, (D) Q]
KTK

Mathematics
Lecture -
07
Sir

1. If x < 0 then cot–1 x in terms of arc sin is
2. If x < 0 then cos–1 x in terms of arc tan is
3. If x > 1 sec–1 x = sin–1
= tan–1

4. If α = β ⇒ sin α =
sin β
(T/F)
5. If sin α = sin β ⇒ α
= β
(T/F)
6. tan–1 x + tan–1 y = ቐ
if
if
7. tan–1 x – tan–1 y = if

Let S = x: cos−1 x = π + sin−1 x + sin−1
2x + 1
. Then ෍ 2x − 1 2 is equal to
x∈S
Ans. 5
Jan)]

Let x ∗ y = x2 + y3 and (x ∗ 1) ∗ 1 = x ∗ (1 ∗ 1). Then a value
of 2 sin−1
x4+x2−
2
x4+x2+
2
is
π
4
π
3
π
2
π
6
(June)]
A
B
C
D

+ sin cot−1
x
2 − 1 1/2
=
QUESTION [IIT-JEE 2008 (Paper 1)]
If 0 < x < 1, then 1 + x2 x cos
cot−1 x
Ans. C
x
1 +
x2
x
x 1 +
x2
1 +
x2
A
B
C
D

Let the inverse trigonometric functions take principal values. The number of real
5
solutions of the equation 2 sin−1 x + 3 cos−1 x =
2π
, is
April)]
Ans. 0

Considering the principal values of the inverse trigonometric functions, the sum of all the
solutions of the equation cos−1 (x) − 2sin−1 (x) = cos−1 (2x) is equal to:
QUESTION [JEE Mains
2022]
A. 0
B. 1
C. 1/2
D. –1/2

4
For n ∈ N, if cot−1 3 + cot−1 4 + cot−1 5 + cot−1 n = π
, then n is
equal to
April)]
Ans. 47

Let S be the set of all solutions of the equation
cos−1 2x − 2 cos−1 1 − x2 = π, x ∈
− ,
1
1
2
2
. Then ෍ 2 sin
x∈S
−
1
x2 − 1 is equal to
π − 2
sin−1
3
4
π −
sin−1
3
4
−2
π3
None
Feb)]
A
B
C
D

Find the number of solution(s) of the equation
(i) cos−11 − x − 2 cos−1 x =
π
(ii) cos−11 − x
2
− 2 cos−1 x
= 1.7
QUESTIO
N

tan−1 x + 1
+ tan−1 x − 1
=
(−7) x − 1 x
QUESTIO
N

Considering only the principal values of inverse trigonometric functions, the number of
4
positive real values of x satisfying tan−1 x + tan−1 2x = π
is :
Jan)]
Ans. D
more than 2
2
0
1
A
B
C
D

The sum of possible values of x for tan−1 x + 1 +
cot−1
1
x−
1
=
tan−1
8
3
1
is:
March)]
Ans. A
3
2
−
4
3
3
−
4
3
1
−
4
3
0
−
4
A
B
C
D

cot−
1
x2 −
1 2
x
+
tan−1
2
x
x2 −
1
=
2
π
3
QUESTIO
N

If the sum of all the solutions of tan−1
2
x
1−x2
+
cot−1
1−x2
2
x
π
= , −1 < x < 1, x ≠
0, is
3
3
α − 4
, then α is equal
to
Jan)]
Ans. 2

−
1
For any y ∈ ℝ, let cot−1 y ∈ (0, π) and tan (y)
∈ − ,
π
π
2
2
. Then the sum of all the
solutions of the equation tan−1 +
cot−1
6y
9−y2
9−y2
6y
3
= 2π
for 0 < |y| < 3, is equal
to:
2)]
Ans. C
2 3 −
3
3 − 2
3
4 3 −
6
6 − 4
3
A
B
C
D

If α > β > γ > 0, then the
expression
cot−1 β
+
+ cot−1 γ
+
1 + β2 1 +
γ2
α − β β −
γ
+ cot−1 α
+
1 +
α2
γ −
α
is equal to :
3
π
2
π
− α + β +
γ
A
B
C
π
D 0
Ans. C
Jan)]

n
Let x = m
(m, n are co-prime natural numbers) be a solution of the equation
cos 2 sin−1 x = 1
and let α, β(α > β) be the roots of the
equation
9
mx2 − nx − m + n = 0. Then the point (α, β) lies on the
line
Jan)]
Ans. C
3x – 2y = –2
3x + 2y = 2
5x + 8y = 9
5x – 8y = –9
A
B
C
D

Mathematics
Lecture -
08
Sir

1. sin−1 x =
ቈ
tan−1
tan
−
1
if x >
0
if x <
0
2. tan−1 x =
ቈ
sin−1
sin−
1
if x >
0
if x <
0
3. cos−1x = ቈcot−1
cot−1
if x >
0
if x <
0
4.
cot−1 x = ቈcos−1
cos−1
if x >
0
if x <
0

5. Expression Substitution
a2 −
x2
a2 −
x2
a−
x
a+
x
x2
6. If θ ∈ −
,
π
π
2
2
then cos θ, sec θ
are
If θ ∈ [0, π] then sin θ, cosec θ are

The value of tan−1 1+x2+ 1−x2
1+x2− 1−x2
1
, |x| < , x ≠ 0, is equal
to
2
π
− 1
cos
x
4 2
−1
2
π
1
4
2
+ cos
x
−1
2
4
π
−
cos−1x2
4
π
+
cos−1x2
QUESTIO
N
A
B
C
D

If S
=
x ∈ ℝ ∶
sin−1
x +
1
x2 + 2x +
2
−
sin−1
x
x2 +
1
4
π
= , then
෍ sin x2 + x + 5
π
x∈s
2
− cos x2 + x + 5
π
is equal to
April)]
Ans. 4

Considering the principal values of the inverse trigonometric functions,
sin−
1
3
x +
1
2 2
1 − x2 ,
−
1
2 2
< x < 1
, is equal to
6
−5π
− sin−1
x
6
5π
− sin−1
x
6
4
π
+ sin−1
x
A
B
C
π
+ sin−1
x
D
Ans. C
April)]

If π
≤ x ≤ 3π
, then cos−1 12
cos x
+
2 4 13
1
3
5
sin
x
is equal to
x + tan−1
5
1
2
x − tan−1
4
3
x + tan−1
4
5
x − tan−1
5
1
2
A
B
C
D
Ans. D
Jan)]

If f(x) = 2tan−1x +
sin−1
2
x
1+x2
, x > 1 then f(5) is equal
to:
/
2

4 tan–1(5)
tan−
1
6
5
15
6
QUESTIO
N
A
B
C
D

Let tan−1y = tan−1x +
tan−1
2
x
1−x2 3
, where |x| < 1
. Then a value of y is:
3x −
x3
1 +
3x2
3x +
x3
1 +
3x2
3x −
x3
1 −
3x2
3x +
x3
1 −
3x2
QUESTIO
N
A
B
C
D

Write the following functions in the simplest form:
x
1+x2−1
, x ≠
0
1
x2−1
, x >
1
5.
tan−1
7.
tan−1
1−cos
x
1+cos
x
, 0 < x <
π
6.
tan−1
8.
tan−1
cos x−sin
x
cos x+sin
x
,
−π
< x <
3π
4
4
QUESTION

The number of solution of the equations 2 sin−1
2
x
1+x2
− πx3 = 0 is equal
to
QUESTIO
N
0
1
2
3
A
B
C
D

Let tan−1 x ∈ −
,
π
π
2
2
, for x ∈ ℝ. Then the number of real solutions of the equation
1 + cos2x = 2 tan−1tan x in the set 2 2 2
2
−
3π
, −
π π π π
,
3π
2
2
∪ − , ∪ is equal to:
1)]
Ans. 3

The number of solutions of the equation tan−1
x
= x2 + 1 2 − 4x2
is
QUESTIO
N

Considering only the principal values of the inverse trigonometric functions, the value of
2
3
cos−1
2 1
2+π2 + 4
sin
−1
2
2
π
2+π2 +
tan
−
1
2
π
is :
QUESTION [JEE Advance
2022]

Let x = sin 2 tan−1 α and y =
sin 1
tan−1 4
.
2 3
If S
=
α ∈ R: y2 = 1 − x , then ෍ 16α3 is
equal to
𝝰∈S
Ans. 130
QUESTION [JEE Mains
2022]

Considering only the principal value of inverse functions, the set
A = x ≥ 0: tan−1 2x +
tan−1 3x =
π
4
contains two elements
contains more than two elements
is a singleton
is an empty set
A
B
C
D
Ans. C
(Jan)]

Number of integers in range of f
x
= x x + 2 x + 4 x + 6 + 7, x ∈ [−4,
2] is
Mind Bender

n=
1
1. The sum σ∞ tan−
1
4
n
n4−2n2+
2
(A) tan−1 1
+ tan−1
2
2
3
(C) π
2
is equal to
(B) 4 tan−1 1
(D) sec−1
−
2
r=
1
σn tan−
1
2r−
1
r2−r+1
−2r3
= 961 then the value of n is equal
to
2. If tan
(A) 31
r2+r+1
(B) 30 (C) 60 (D) 61
3. If the solution set of inequality cosec−1x 2 − 2 cosec−1x
≥ π
6
(−∞, m] ∪ [n, ∞) then (m + n) equals
(A) 0 (B) 1 (C) 2 (D) –3
QUESTIO
N
[Ans. D]
[Ans. A]
cosec−1x − 2 is
[Ans. B]

If a sin−1 x − b cos−1 x = c, then the value of a sin−1 x + b cos−1 x (whenever
exists) is equal to:
0
πab + c
b −
a
a +
b
π
2
πab + c
a −
b
a +
b
A
B
C
D
Ans. D
QUESTIO
N

The range of f
x
=
cot−1
−x − tan−1 x + sec−1 x
is :
−
,
π
3π
2
2
π
3π
, π ∪ π,
2 2
−
,
π
3π
2
2
π
3π
, π ∪ π,
2 2
A
B
C
D
Ans. B
QUESTIO
N

Mathematics
Lecture -
09
Sir

1. cos–1 (–x) =
sin–1 (–x) =
tan–1 (–x) =
sec–1 (–x) =
cot–1 (–x) =
cosec–1 (–x) =
if
2. tan–1 x + tan–1 y = ቐ
if

3. If x, y > 0,
tan−1
x−
y
1+x
y
=
4. If x > 0,
tan−1
x
1−6x2
=
5. =
3
෍
ta
n−
1
r=1
2
r2

6. sin–1 x is function cos–1 x is
function tan–1 x is
function cot–1 x is function
7. [sin–1 x] can take value =
8. [cos–1 x] can take value =

Considering only the principal values of the inverse trigonometric functions, the value of
2
3
cos−1
2 1
2+π2 + 4
sin
−1
2
2
π
2+π2 +
tan
−
1
2
π
is :
QUESTION [JEE Advanced
2022]

Solve the inequality: sin−1 x > sin−1(3x −
1)
QUESTION

sin−1 x > cos−1 x & sin−1 x > sin−1(1
− x)
QUESTION

Solve the system of inequalities involving inverse circular functions
arc tan2 x − 3 arc tan x + 2 > 0 and sin−1 x >
cos−1 x
where [.] denotes the greatest integer function.
QUESTION

cot−1 x 2 − 6 cot−1 x + 9 ≤ 0, where [.]
denotes the
Find the complete solution set of
greatest integer function.
QUESTION

Solve the following system of in equations
4 arc tan2 x − 8 arc tan x + 3 < 0
and 4 arc cot x − arc cot2x − 3
> 0
QUESTION

tan−
1
3
2
π
2
cot−
1
3
2
tan−1(3
)
k
Let Sk = ෍ tan−1
r=1
6r
22r+1 +
32r+1
k→
∞
then lim Sk is equal
to:
QUESTION [JEE Mains
2021]
A
B
C
D

101/102
50/51
100
51/50
50
If ෍ tan−1
r=1
1
2r2
= p, then the value of tan p is:
QUESTION [JEE Mains
2021]
A
B
C
D

If cot–1 () = cot–1 2 + cot–1 8 + cot–1 18 + cot–1 32 +….. upto 100 terms, then  is
QUESTION [JEE Mains
2019]
A 1.01
B 1.00
C 1.02
D 1.03

n
෍
ta
n−
1
r=1
2r−1
1 +
22r−1
is equal
to
tan−
1
2n
tan−
1
2n
π
−
4
tan−
1
2n+1
tan−
1
2n+1
π
−
4
QUESTIO
N
A
B
C
D

50
The value of cot ෍ tan−1
n=1
1
1 + n +
n2
is
2
6
2
5
2
5
2
6
5
0
5
1
5
2
5
1
QUESTION [JEE Mains 2022 (June)]
A
B
C
D

5/11
If S is the sum of the first 10 terms of the series
tan−
1
1
3
+
tan−1
+ tan−1 +
tan−1
1 1
1
7
13
21
+
QUESTION [JEE Mains
2020]
then tan(S) is equal to:
A 10/11
B –6/5
C 5/6
D

If y = cos π
+ cos−1 x
, then (x − y)2 + 3y2 is
equal to
3 2
Ans. 3
April)]

The sum of the infinite series cot−1
7
4
+
cot−1
1
9
4
+
cot−1
3
9
4
+
cot−1
6
7
4
+ ⋯
is:
2
π
+
cot−1
1
2
2
π
−
cot−1
1
2
2
π
−
tan−1
1
2
2
π
+
tan−1
1
2
A
B
C
D
Ans. C
April)]

Let a, b ∈ (0, 2π) be the largest interval for which sin−1 sin θ − cos−1 sin
θ > 0, θ ∈ (0, 2π) holds. If αx2 + βx + sin−1 x2 − 6x + 10 + cos−1 x2 −
6x + 10 = 0 and α − β = b − a, then α is equal to:
π
4
8
π
1
6
π
8
π
1
2
A
B
C
D
Ans. D
QUESTIO
N

0, π
is defined by g(x) =
cos−1
x2−k
3
1+x2
.
Then the possible values of 'k' for which g is surjective function, is
1
2 2
(A) (B) −1, − 1
(C) −
1
2
1
2
(D) − ,
1
2. Number of values of x satisfying the equation cos 3 arc cos x
− 1
(A) 0 (B) 1 (C) 2
(D) 3
3. The number of ordered triplets
4
2
sin−1 x 2 =
π
+
(A) 2
sec−1 y 2
+
(B) 4
x, y, z satisfy the equation
tan−1 z 2 is
(C) 6
(D) 8
QUESTION
1. Let g: R
→
KTK
[Ans. A]
[Ans. C]
= 0 is equal to
[Ans. D]

The number of solutions of the equation |y| = cos x and y =
cot−1
cot x in −
3π
,
5π
2
2
is
2
4
6
none of these
A
B
C
D
Ans. A
QUESTIO
N

Let f: R → 0,
π
6
be defined as f(x) =
sin−1
4
4x2−12x+1
7
then f(x)
is :
injective as well as surjective
surjective but not injective
injective but not surjective
neither injective nor surjective
A
B
C
D
Ans. B
QUESTIO
N

Mathematics
Lecture -
10
Sir

If
cos−1
2
3
x
+
cos−1
3
=
π 4x
2 4
x > 3
then x is equal to:
14
5
1
2
14
5
1
0
14
6
1
2
14
5
1
1
QUESTION [JEE Mains
2019]
A
B
C
D

The value of sin−1
1
2
1
3
−
sin−1
3
5
is
π −
cos−1
3
3
6
5
π −
sin−1
6
3
6
5
2
π
−
sin−1
5
6
6
5
2
π
−
cos−1
9
6
5
QUESTION [JEE Mains
2019]
A
B
C
D

Given that the inverse trigonometric function assumes principal values only. Let x, y be
2
any two real numbers in [−1, 1] such that cos−1 x − sin−1 y = α, −π
≤ α ≤ π.
Then, the
minimum value of x2 + y2 + 2xy sin
α is
April)]
Ans. A
0
–1
1/2
–1/2
A
B
C
D

If cos−1x − cos−1 y
= α, where −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, x ≤ y
, then
for all x, y,
QUESTION [JEE Mains
2019]
2
2
4x2 − 4 xy cosα + y2 is equal to
A. 2 sin2α
B. 4 cos2α + 2x2y2
C. 4 sin2α
D. 4 sin2α − 2x2y2

Let (x, y) be such that sin−1 ax + cos−1 y + cos−1
bxy
=
π
2
Match the statements in Column I with statements in Column II.
Ans. (A) P; (B) Q; (C) P; (D) S
QUESTION [IIT-JEE
2007]
Column I Column II
(A) If a = 1 and b = 0, thenx, y (P) lies on the circle x2 + y2 = 1
(B) If a = 1 and b = 1, thenx, y (Q) lies on x2 − 1 y2 − 1 = 0
(C) If a = 1 and b = 2, thenx, y (R) lies on y = x
(D) If a = 2 and b = 2, thenx, y (S) lies on 4x2 − 1 y2 − 1 = 0

The number of solutions of the equationsin−1 x2 + 1
+
cos−1
3
3
x2 − 2
= x2
for
x ∈ −1, 1and [x] denotes the greatest integer less than or equal to x, is
0
infinite
2
4
A
B
C
D
Ans. A
QUESTION [JEE Mains
2021]

The complete solution set of the equation
sin−
1
2
1+x
−
2 − x = cot−1 tan
2 − x
−
sin−1
2
1−x
is
[–1, 1]
4
2
2 − π
,
1
2
−1,2 −
π 4
[0, 1]
A
B
C
D
Ans. B
QUESTIO
N

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