Chapter
2
Inverse Trigonometric
Functions
Domain, Range and Principal Value of Inverse Circular Functions
Function
x
[–1, 1]
Range
− π , π 
 2 2 
Principal value branch
π
π
− ≤ y ≤ , where
2
2
y = sin −1 x
cos−1 x
[–1, 1]
[0, π]
0 ≤ y ≤ π, where y = cos−1 x
tan −1 x
R
− π, π


 2 2
cosec −1 x
(−∞, − 1 ] ∪ [1, ∞)
 − π , π  − {0 }
 2 2 
sec −1 x
(−∞, − 1 ] ∪ [1, ∞)
π
[0, π] −  
2
sin
−1
Domain
π
π
< y< ,
2
2
where y = tan −1 x
−
π
π
≤ y ≤ , y ≠ 0,.
2
2
where y = cosec −1 x
−
π
0 ≤ y ≤ π, y ≠  ,
2
where y = sec −1 x
cot
−1
(0, π)
R
x
0 < y < π, where y = cot −1 x
Properties of Inverse Circular Functions
1. sin −1(sin θ) = θ
2. cos −1(cos θ) = θ
3. tan −1(tan θ) = θ
4. cot −1(cot θ) = θ
−1
6. cosec −1(cosec θ) = θ
 1
8. cos −1   = sec −1x
 x
5. sec (sec θ) = θ
 1
7. sin −1   = cosec −1x
 x
 1
9. tan −1   = cot −1 x
 x
10. sin −1(− x) = − sin −1 x
11. cos −1(− x) = π − cos −1 x
−1
13. cot (− x) = π − cot
−1
x
12. tan −1(− x) = − tan −1 x
14. sec −1(− x) = π − sec −1x
44
www.arihantbooks.com
15. cosec −1(− x) = − cosec −1x
17. tan −1 x + cot −1 x =
π
2
16. sin −1 x + cos −1 x =
π
2
18. sec −1 x + cosec −1x =
π
2
19. sin −1 x + sin −1 y = sin −1[x 1 − y 2 + y 1 − x 2 ]
20. sin −1 x − sin −1 y = sin −1 [x 1 − y 2 − y 1 − x 2 ]
21. cos −1 x + cos −1 y = cos −1 [xy − 1 − x 2
1 − y2]
22. cos −1 x − cos −1 y = cos −1 [xy + 1 − x 2
1 − y2]
 x + y
23. tan −1 x + tan −1 y = tan −1 
 , x > 0, y > 0 and xy < 1
 1 − xy 
 x + y
24. tan −1 x + tan −1 y = π + tan −1 
 , x > 0, y > 0 and xy > 1
 1 − xy 
 x + y
25. tan −1 x + tan −1 y = tan −1 
 − π, x < 0, y < 0 and xy > 1
 1 − xy 
 x−y
26. tan −1 x − tan −1 y = tan −1 
 , xy > − 1
 1 + xy 
 x−y
27. tan −1 x − tan −1 y = π + tan −1 
 , x > 0, y < 0 and xy < − 1
 1 + xy 
 x−y
28. tan −1 x − tan −1 y = tan −1 
 − π, x < 0, y > 0 and xy < − 1
 1 + xy 
2
 2x 
−1  2 x 
−1  1 − x 
29. 2 tan −1 x = tan −1 
sin
cos
=
=





1 − x2
1 + x2
1 + x2
30. 2 sin −1 x = sin −1(2x 1 − x 2 )
31. 2 cos −1 x = cos −1(2x 2 − 1)
32. 3 sin −1 x = sin −1(3x − 4x 3)
33. 3 cos −1 x = cos −1(4x 3 − 3x)
 3x − x 3 
34. 3 tan −1 x = tan −1 

 1 − 3x 2 

x 
35. sin −1 x = cos −1 1 − x 2 = tan −1 
 1 − x2 


 1 − x2 

36. cos −1 x = sin −1 1 − x 2 = tan −1 

x 




x
1 
 = cos −1 
37. tan −1 x = sin −1 
 1 + x2 
 1 + x2 




NCERT Class XII Mathematics Solutions
45
Exercise 2.1
Direction (Q. Nos. 1 to 10) Find the principal values of the following
questions :
 3
 1
2. cos −1 
3. cosec −1 (2)
4. tan −1 (− 3)
1. sin −1  − 

 2
 2 
 2 
 1
−1
5. cos −1  − 
6. tan −1 (−1)
7. sec −1 
 8. cot ( 3)
 2
 3
 1
−1
9. cos −1  −
 10. cosec (− 2)

2
Solutions for 1-10
1
 1
1. Let sin −1  −  = θ ⇒ sin θ = −
 2
2
 π π
We know that the range of principal value of sin −1 θ is − , .
 2 2 
1
π
 π
(Qsin (− θ) = − sin θ)
∴ sin θ = − = − sin = sin  − 
 6
2
6
π
π
 π π
 1
⇒
θ = − , where θ ∈ − ,
⇒ sin −1  −  = −
 2
6
6
 2 2 
π
 1
Hence, the principal value of sin −1  −  is − .
 2
6
Note Principal value of any inverse function is unique.
 3
3
2. Let cos −1 
 = θ ⇒ cos θ =
2
 2 
We know that the range of principal value of cos −1 θ is [0, π ].
 3 π
3
π
π
⇒ θ = , where θ ∈[0, π] ⇒ cos −1 
= cos
∴ cos θ =
 =
2
6
6
 2  6
 3 π
Hence, the principal value of cos −1 
 is .
 2  6
3. Let cosec −1 2 = θ ⇒ cosec θ = 2. We know that the range of principal
π
 π π
value of cosec −1 θ is − ,
− {0}. cosec θ = 2 = cosec
 2 2 
6
π
π
 π π
−1
⇒
θ = , where θ ∈ − ,
− {0} ⇒ cosec (2) =


6
6
 2 2
π
Hence, the principal value of cosec −1(2) is .
6
4. Let tan −1 (− 3) = θ ⇒ tan θ = − 3
 π π
We know that the range of principal value of tan −1 θ is  − ,  .
 2 2
46
www.arihantbooks.com
π
 π
(Qtan (− θ) = − tan θ)
= tan  − 
 3
3
π
π
 π π
⇒
θ = − , where θ ∈  − ,  ⇒ tan −1 (− 3) = −
 2 2
3
3
π
−1
Hence, principal value of tan (− 3) is − .
3
∴
tan θ = − 3 = − tan
1
 1
5. Let cos −1  −  = θ ⇒ cos θ = −
 2
2
We know that the range of principal value of cos −1 θ is [0, π].
1
π
π

(Qcos (π − θ) = − cos θ)
∴ cos θ = − = − cos = cos  π − 

2
3
3
2π
= cos
3
2π
 1 2π
; where θ ∈[0, π] ⇒ cos −1  −  =
θ=
⇒
 2
3
3
 1 2π
.
Hence, principal value of cos −1  −  is
 2
3
Note cos −1 (− θ) ≠ − cos −1 θ
6. Let tan −1 (−1) = θ ⇒ tan θ = − 1
 π π
We know that the range of principal value of tan −1 θ is  − ,  .
 2 2
π
 π
(Qtan (− θ) = − tan θ)
∴
tan θ = − 1 = − tan = tan  − 
 4
4
π
π
 π π
∴ tan −1(−1) = −
θ = − ; where
θ ∈− , 
⇒
 2 2
4
4
π
Hence, the principal value of tan −1 (−1) is − .
4
2
−1  2 
7. Let sec 
 = θ ⇒ sec θ =
 3
3
π 
We know that the range of principal value of sec −1 θ is [0, π ] −  .
2 
2
π
π
π 
∴
sec θ =
= sec
⇒ θ = , where θ ∈ [0, π ] −  
6
6
3
2 
π
−1  2 
⇒
sec 
 =
 3 6
 2  π
Hence, the principal value of sec −1 
 is .
 3 6
NCERT Class XII Mathematics Solutions
47
8. Let cot −1 ( 3) = θ ⇒ cot θ = 3
We know that the range of principal value of cot −1 θ is (0, π ).
π
π
π
∴ cot θ = 3 = cot ⇒ θ = ; where θ ∈(0 , π ) ⇒ cot −1 ( 3) =
6
6
6
π
Hence, the principal value of cot −1 ( 3) is .
6
1
1
−1 
9. Let cos  −
 = θ ⇒ cos θ = −

2
2
We know that the range of principal value of cos −1 θ is [0, π].
1
π
π

[Qcos(π − θ) = − cos θ)]
∴
cos θ = −
= − cos = cos  π − 

4
4
2
3π
 1  3π
, where θ ∈[0, π] ⇒ cos −1  −
θ=
⇒
 =

4
4
2
1  3π
−1 
.
Hence, the principal value of cos  −
 is

4
2
10. Let cosec −1 (− 2) = θ
We know that the range of principal value of cosec −1 θ is
 π π
− ,
− {0}.
 2 2 
π
 π
∴ cosec θ = − 2 = − cosec = cosec  −  (Qcosec (−θ) = − cosec θ)
 4
4
π
π
 π π
⇒
− {0} ⇒ cosec −1(− 2) = −
θ = − , where θ ∈ − ,
 2 2 
4
4
π
−1
Hence, the principal value of cosec (− 2) is − .
4
Direction (Q. Nos. 11 to 14) Find the values of the following questions :
1
1
Question 11. tan −1 (1) + cos −1  −  + sin −1  − 

2

2
Given expression is not a standard identity, so we separately find the
1
1
value of tan−1 ( 1), cos −1  −  , sin−1  −  and simplify it.
 2
 2
Solution Let tan −1(1 ) = x ⇒ tan x = 1 = tan
π
π
⇒x =
4
4
π
 π π
where principal value x ∈  − , 
∴ tan −1 (1) =
 2 2
4
1
π
 π

 2π 
−1  1
Let cos  −  = y ⇒ cos y = − = − cos   = cos  π −  = cos  
 2
 3

 3
2
3
(Qcos (π − θ) = − cos θ)
2π
, where principal value y ∈[0, π ]
⇒
y=
3
48
www.arihantbooks.com
∴
 1 2π
cos −1  −  =
 2
3
Let
⇒
∴
∴
1
 1
 π
 π
sin −1  −  = z ⇒ sin z = − = − sin   = sin  − 
 2
 6
 6
2
π
 π π
z = − , where principal value z ∈ − ,
 2 2 
6
π
 1
sin −1  −  = −
 2
6
π 2π π
 1
 1
tan −1 (1) + cos −1  −  + sin −1  −  = x + y + z = +
−
 2
 2
4
3
6
3π + 8π − 2π 9π 3π
=
=
=
12
12
4
1
1
Question 12. cos −1   + 2 sin −1  
 2
 2
Solution We can find the value of given expression by simplifying the
individual terms.
1
π
 1
cos −1   = x ⇒ cos x = = cos
 2
2
3
π
(principal interval)
x = ∈ [0, π]
⇒
3
1
π
 1
Again, let
sin −1   = y ⇒ sin y = = sin
 2
2
6
π  π π
(principal interval)
y= ∈ − ,
⇒
6  2 2 
π
π 2π
 1
 1
cos −1   + 2 sin −1   = x + 2y = + 2 × =
∴
 2
 2
3
6
3
Let
Question 13. If sin −1 x = y, then
(a) 0 ≤ y ≤ π
(b) −
π
π
≤ y≤
2
2
π
π
< y<
2
2
π
π
π π
, therefore − ≤ y ≤ .
,
2
2
2 2 
(c) 0 < y < π
Solution (b) As range of sin −1 x is −

(d) −
Hence, the correct option is (b).
Question 14. tan −1 3 − sec−1 ( −2) is equal to
(a) π
Solution
⇒
2π
3
π
(b) Let tan −1 3 = x ⇒ tan x = 3 ⇒ tan x = tan
3
π  π π
(principal interval)
x = ∈− , 
3  2 2
(b) −
π
3
(c)
π
3
(d)
NCERT Class XII Mathematics Solutions
Let
sec −1 (−2) = y
⇒ sec y = − 2
sec y = − sec
⇒
49
π
π

⇒ sec y = sec  π − 

3
3
[Qsec (π − θ) = − sec θ]
2π
 2π 
 π
sec y = sec   ⇒ y =
∈ [0, π] −  
 3
 2
3
π 2π
π
∴
tan −1 3 − sec −1(−2) = x − y = −
=−
3
3
3
Hence, the correct option is (b).
⇒
(principal interval)
Exercise 2.2
Direction (Q. Nos. 1 to 4)
Prove the following questions :
Question 1. 3 sin −1 x = sin −1 ( 3 x − 4 x3 ), x ∈ − , 
 2 2
1 1
The right hand side of a given expression can be converted into a
standard formula sin 3θ when we substitute sin−1 x as θ and then applying
the formula sin 3 θ = 3 sin θ − 4 sin3 θ.
Solution Let sin −1 x = θ ⇒ x = sin θ, then
LHS = 3 sin −1 x = 3 sin −1 (sin θ) = 3 θ
RHS = sin −1 (3x − 4x 3) = sin −1 [3 sin θ − 4 sin 3 θ] = sin −1 [sin 3θ] = 3 θ
[Qsin 3 θ = 3 sin θ − 4 sin 3 θ]
LHS = RHS,
Hence proved.
∴
1 
−1
−1
3
Question 2. 3 cos x = cos (4 x − 3 x), x ∈  , 1.
2 
The right hand side of a given expression can be converted into a
standard formula cos 3θ when we substitute cos −1 x as θ and then
applying the formula cos 3 θ = 4 cos 3 θ − 3 cos θ.
Solution Let x = cos θ
LHS = 3 cos −1 x = 3 cos −1 (cos θ) = 3 θ
RHS = cos −1 (4x 3 − 3x) = cos −1 [4 cos 3 θ − 3 cos θ] = cos −1 [cos 3θ] = 3 θ
[Qcos 3 θ = 4 cos 3 θ − 3 cos θ]
∴
LHS = RHS,
2
7
1
+ tan −1
= tan −1
11
24
2
1
−1 2
−1 7
Given, tan
+ tan
= tan −1
11
24
2
Question 3. tan −1
Solution
Hence proved.
50
www.arihantbooks.com
7 
 2
+


2
7




LHS = tan −1   + tan −1   = tan −1  11 24 
 11
 24
1 − 2 . 7 

11 24 

 x + y 
Q tan −1 x + tan −1 y = tan −1 

 1 − xy  

 48 + 77
 125 




−1
−1
264
= tan 
 = tan  264  = tan
−
14
264
14
1−




 264 
264 
−1
 125 264
−1 1
= RHS.
= tan −1 
×
 = tan
 264 250
2
125
264
250
264
Hence proved.
1
1
31
Question 4. 2 tan −1   + tan −1   = tan −1  
 2
 7
 17 
 2x 
 so that we can applying
Firstly we use the identity 2 tan−1 x = tan−1 
2
1 − x 
 A+ B 
tan−1 ( A) + tan−1 ( B) = tan−1 
 to get the required result.
 1 − AB 
1
1
31
Solution Given, 2 tan −1   + tan −1   = tan −1  
 
 
 
2
7
17


 2×1 
 1
 1
2  + tan −1  1
LHS = 2 tan −1   + tan −1   = tan −1 
2

 7
 2
 7
1
  
1 −   
 2 

= tan −1
1

 2x  
Q2 tan −1 x = tan −1 

1 − x2

1
 4
 1
= tan −1   + tan −1  
 3
 7
7
+ tan −1
1
1−
4
 4 1 
+



 x + y 
= tan −1  3 7 
Q tan −1 x + tan −1 y = tan −1 

4
1
 1 − xy  

1 − × 

3 7
 28 + 3 
 31


 
21
 31
−1  31
−1
−1 21
21
= tan 
×  = tan −1   = RHS
 = tan   = tan 
4
17
 17


21 17
 1−

 

 21 
21 
Hence proved.
NCERT Class XII Mathematics Solutions
51
Direction (Q. Nos. 5 to 10) Write the following functions in the simplest
form.
Question 5. tan −1
1 + x2 − 1
,x≠0
x
To remove the square root sign, we substituting x as tan θ and simplify it.
Solution Let x = tan θ, then θ = tan −1 x
∴
tan −1
1 + tan θ − 1
1+ x − 1
= tan −1
= tan −1
tan θ
x
2
2
…(i)
sec θ − 1
tan θ
[Q1 + tan 2 θ = sec 2 θ]
2
 sec θ − 1
= tan −1 

 tan θ 

 1
 1 − cos θ 
−1


cos θ 
−1 cos θ
−1  cos θ 
−1  1 − cos θ
= tan 
 = tan 
 = tan  cos θ × sin θ 
sin
in
θ
θ
s






 cos θ 
 cos θ 
θ 
θ 


2 sin 2
Q1 − cos θ = 2 sin 2



−1  1 − cos θ 
−1 
2
2
= tan 
= tan

 sin θ 
θ
θ
θ
θ
2 sin cos 
and sin θ = 2 sin cos 

2
2

2
2
θ

2 sin
−1

2  = tan −1 tan θ  = θ = tan x
[from Eq. (i)]
= tan −1 

θ
2  2
2
2 cos 

2
tan −1
∴
Question 6. tan −1
1
x −1
2
1 + x2 − 1 1
= tan −1 x
x
2
, |x| > 1
Substitute x as sec θ to remove the square root sign and simplify it.
Solution Let x = sec θ, then θ = sec −1 x
∴
tan −1
…(i)



1
1 
 = tan −1 
= tan −1 

 sec 2 θ − 1
 tan 2 θ 
x2 − 1


1
(Qsec 2 θ − tan 2 θ = 1)
 1 
π


π
−1
−1 
= tan −1 
 = tan (cot θ) = tan  tan  − θ  (Qtan  − θ = cot θ)

2
 tan θ 
2


52
www.arihantbooks.com
=
π
π
− θ = − sec −1 x
2
2
[from Eq. (i)]
 1 − cos x 
, x < π
 1 + cos x 
Question 7. tan −1 
Applying the formula 1 − cos x = 2sin2( x / 2) and 1 + cos x = 2cos 2( x / 2) to
remove the square root sign and simplify it.
 2 sin 2(x / 2) 
 1 − cos x 

 = tan −1 
2

 1 + cos x 
 2 cos (x / 2) 
Solution tan −1 

1 − cos x = 2 sin 2(x / 2) 

Q
 and 1 + cos x = 2 cos 2(x / 2)

x
x x

= tan −1  tan 2  = tan −1  tan  =

2
2 2

Note If x is positive, then
tan 2
tan 2
x
x
= tan and if x is negative, then
2
2
x
x
= − tan .
2
2
 cos x − sin x 
, 0 < x < π
 cos x + sin x
Question 8. tan −1 
First we convert the angle of given function in tan θ so that we use the
identity. For this we have change the angle of numerator and
denominator in terms of tan x and then using the formula
tan a − tan b
tan( a − b) =
1 + tan a tan b
 cos x sin x 
−


x − sin x 
−1 cos x
cos x 
Solution tan 
 = tan 
cos
x
sin
x
 cos x + sin x 


+
 cos x cos x 
(inside the bracket divide numerator and denominator by cos x)
 1 − tan x 

π
−1 
= tan −1 
 = tan tan  − x 


 1 + tan x 
4

−1  cos
π


tan − tan x

1 − tan x 
π

4
=
Q tan  4 − x  =

π
1 + tan ⋅ tan x 1 + tan x 



4
=
π
−x
4
NCERT Class XII Mathematics Solutions
Question 9.
x
tan −1
a 2 − x2
53
, |x| < a
Substitute x as a sin θ to remove the sqrare root sign and then use the
identity.
Solution Let x = a sin θ ⇒
∴
tan −1
x
a2 − x 2
x
 x
= sin θ ⇒ sin −1   = θ
 a
a
…(i)


a sin θ

= tan −1 
 a2 − a2 sin 2 θ 


 a sin θ 
 = tan −1  sin θ 
= tan −1 
 a 1 − sin 2 θ 
 cos θ 



sin 2 x + cos 2 x = 1 
Q

2
 ⇒ cos x = 1 − sin x 
 x
= tan −1(tan θ) = θ = sin −1  
 a
[from Eq. (i)]
 3 a 2 x − x3 
−a
a
≤x≤
 , a > 0;
3
3
 a 3 − 3 ax2 
Question 10. tan −1 
The angle of given function can be converted into standard result of
tan 3 θ by putting x = a tan θ, so that we use the identity tan−1 (tan θ) = θ,
3tan θ − tan3 θ
where, tan 3 θ =
1 − 3tan2 θ
Solution Let x = a tan θ
⇒
∴
x
 x
= tan θ ⇒ θ = tan −1  
 a
a
… (i)
 3a2x − x 3 
 3a2(a tan θ) − (a tan θ)3 
tan −1  3
 = tan −1 

2
a3 − 3a(a tan θ)2
 a − 3ax 


 a3(3 tan θ − tan 3 θ)
= tan −1 

 a3(1 − 3 tan 2 θ) 
 3 tan θ − tan 3 θ 
= tan −1 
 = tan −1(tan 3θ)
 1 − 3 tan 2 θ 

3 tan θ − tan 3 θ 
Q tan 3θ =

1 − 3 tan 2 θ 

 x
= 3θ = 3 tan −1  
 a
[from Eq.(i)]
54
www.arihantbooks.com
Direction (Q. Nos. 11 to 15) Find the value of each of the following
questions :
1 

Question 11. tan −1 2 cos  2 sin −1  .

2 



1 

π 
Solution tan −1 2 cos 2 sin −1   = tan −1 2 cos 2 sin −1  sin  


2 
6 




π 1

Q sin = 

6 2
π 
π



= tan −1 2 cos 2 ×   = tan −1 2 cos


6 
3 

1
π 1


= tan −1 2 ×
= tan −1(1)
Qcos = 

2 
3 2

π
π
π




= tan −1  tan  =
Q tan = 1



4 4
4
Question 12. cot (tan −1 a + cot −1 a).
π
π

−1
−1
=0
Q tan x + cot x = 

2
2
2

1
2x
1− y
+ cos −1
tan sin −1
 , |x| < 1, y > 0 and xy < 1
2
1 + x2
1 + y2 
Solution cot (tan −1 a + cot −1 a) = cot
Question 13.
The angle of tan is converted into a tan−1 x by using the standard relation
so that we use the identity tan (tan−1 θ) = θ. For this we have to use the
 2x 
 1 − x2 
 = cos −1 

relation 2 tan−1 x = sin−1 
2
2
1 + x 
1 + x 
1
2x
1 − y2 
Solution tan sin −1
+ cos −1

2
1 + x2
1 + y2 

 1 − y2  
2x 
−1
−1 

 and 2 tan −1 y = cos −1 
Q2 tan x = sin 
2
1 + x 
1 + y2 

1

1

= tan (2 tan −1 x + 2 tan −1 y) = tan × 2 (tan −1 x + tan −1 y)
2

2

−1
−1
= tan (tan x + tan y)

 x + y x + y 
x+y
= tan tan −1 
Q tan −1 x + tan −1 y = tan −1
 =

 1 − xy   1 − xy 
1 − xy 

Question 14. If sin sin −1
1

+ cos −1 x = 1, then find the value of x.

5
Solution Given, sin  sin −1
1

+ cos −1 x = 1

5


NCERT Class XII Mathematics Solutions
⇒
sin −1
⇒
sin −1
⇒
sin −1
⇒
⇒
55
1
+ cos −1 x = sin −1 1
5
1
π

+ cos −1 x = sin −1  sin 

5
2
[Qsin θ = x ⇒θ = sin −1 x]
1
π
+ cos −1 x =
⇒ sin −1
5
2
1
sin −1 = sin −1 x
5
1
=x
5
1 π
= − cos −1 x
5 2
π

Q sin −1 x + cos −1 x =

2 


 π
Q sin  2  = 1


x −1
x +1 π
+ tan −1
= , then find the value of x.
x−2
x+ 2 4
x −1
x+1 π
Solution Given tan −1
+ tan −1
=
x −2
x+2 4
 x −1 x + 1 


+
 x + y
x −2 x + 2  π 
−1 
⇒ tan
Q tan −1 x + tan −1 y = tan −1 
=


 1 − xy  
 x − 1  x + 1  4 
1 − 


 x − 2  x + 2 

 (x − 1) (x + 2) + (x + 1)(x − 2) 


(x − 2)(x + 2)
 =π
tan −1 
⇒
 (x − 2) (x + 2) − (x − 1)(x + 1)  4




(x − 2)(x + 2)
(x − 1) (x + 2) + (x + 1) (x − 2)
π
= tan
⇒
(x − 2) (x + 2) − (x − 1) (x + 1)
4
2x 2 − 4
π


⇒
=1
Q tan = 1


4
(x 2 − 4) − (x 2 − 1)
1
1
2
2
2
⇒ x=±
⇒
2x − 4 = − 3 ⇒ 2x = 1 ⇒ x =
2
2
Direction (Q. Nos. 16 to 18) Find the values of each of the expressions.
Question 15. If tan −1
Question 16. sin −1 sin

Since,
sin
the
range
2π 

3
of sin−1 x
is
− π , π 
 2 2 
so,
firstly
2π
π
= sin  π −  and then use the identity sin−1 (sin θ) = θ.

3
3
Solution sin −1  sin

we
write
2π 
π 
π

−1 
−1 
 = sin sin  π −   = sin  sin 



3
3
3


[Q sin (π − θ) = sin θ]
56
www.arihantbooks.com
π
 π π
which lies in the range − , .
 2 2 
3
 π π
Note sin −1(sin θ) = θ is only when θ ∈ − , .
 2 2 
3π
Question 17. tan −1  tan 

4 
π
3
π
π 



Solution tan −1  tan  = tan −1 tan  π −   = tan −1  − tan 




4
4
4


[Qtan (π − θ) = − tan θ]
 π 
−1 
[Q− tan θ = tan (−θ)]
= tan tan  −  
 4 

π
 π π
= − which lies in the range  − ,  .
 2 2
4
=
Question 18. tan sin −1

3
3
+ cot −1 
5
2
The angle of tan be converted into tan−1 so that we can use the identity
3
3
tan (tan−1 θ) = θ, for this we have to convert sin−1 and cot −1 to tan−1 θ
5
2
 a+ b
and applying the formula tan−1 a + tan−1 b = tan−1 
.
 1 − ab 
Solution tan  sin −1

3
+ cot −1
5


3


3
2
−1

5
+ tan −1 
 ⇒ tan tan
2
2
3
 3


1−  
 5



x
x
y
and cot −1 = tan −1 
Q sin −1 x = tan −1
y
x

1 − x2


 3 2 
17


 +  

 4 3
3
2



−1
= tan tan −1 + tan −1
= tan tan −1
 = tan tan 12 


3
2
1
4
3

1− × 



4 3 
2 


−1
−1
−1  x + y  
Q tan x + tan y = tan  1 − xy  


17

−1 17
= tan tan
=

6  6
Question 19. cos −1  cos

(a)
7π
6
(b)
5π
6
7π 
 is equal to
6
π
(c)
3
(d)
π
6
NCERT Class XII Mathematics Solutions
57
7π
according to the interval [ 0, π] i.e., by
6
using cos θ = cos( 2 π − θ) and then use the identity cos −1 (cos θ) = θ.
First we reduce the quantity

7π 
5π  

−1 

 = cos cos 2π −

6
6  

∴
7π 

 5π   5π

cos −1 cos  = cos −1 cos    =
 6  6

6

Solution (b) cos −1 cos
where,
5π
∈[0, π]
6
[Q cos (2π − θ) = cos θ]
Hence, the correct option is (b).
1 
π
Question 20. sin  − sin −1  −   is equal to

3
2 

1
1
1
(b)
(c)
(d) 1
(a)
2
3
4
π 1
1 
π 
π
π
Solution (d) sin  − sin −1  −   = sin  − sin −1  − sin   Q sin = 




6 2
3
2
3
6



 
π
  π  
= sin  − sin −1 sin  −  
  6  
3
(Qsin(− x) = − sin x)
π
π  π 
π π 
= sin = 1
= sin  −  −   = sin
+
 3 6 
2
 3  6 
Hence, the correct option is (d).
Question 21. tan −1 3 − cot −1 ( − 3 ) is equal to
(a) π
(b) −
π
2
(c) zero
(d) 2 3
Solution (b) tan −1 3 − cot −1(− 3)
= tan −1 3 − [π − cot −1 3]
[Qcot −1(− x) = π − cot −1 x]
= tan −1 3 − π + cot −1 3
= tan −1 3 − π + tan −1


−1 1
= cot −1 x
Q tan


x
1
3
1 

= tan −1 3 + tan −1
−π
3 

= tan
−1
1

 3+


3
−π
1

1−  3 ×


3
= tan −1(∞) − π

−1
−1
−1  x + y  
Q tan x + tan y = tan  1 − xy  


58
www.arihantbooks.com
=
π
π
−π=−
2
2
π


Q tan = ∞


2
Hence, the correct option is (b).
Miscellaneous Exercise
Direction (Q. Nos. 1 and 2)
Find the value of following functions :
13 π 
.

6 
13π 
π 
π

−1 
Solution cos −1 cos
 = cos cos 2π +   ; where ∈[0, π]


6 
6 
6

[Hence, the given angle does not lie in the principal interval i.e., [0 π],
so we convert it such that it lies in [0 π].]

 π  π
= cos −1 cos    =
[Q cos(2π + θ) = cos θ]
 6  6

Question 1. cos −1  cos
Question 2. tan −1  tan

7π 

6
π  π π
7π 
π 

−1 
 = tan tan  π +  ; where ∈  − , 


6  2 2
6
6


(Principal interval)
7π 
π π
−1 
−1 
[Qtan (π + θ) = tan θ]
tan  tan  = tan tan
=


6
6  6
Solution tan −1  tan

∴
Direction (Q. Nos. 3 to 12) Prove the following functions :
3
24
= tan −1
5
7
24
−1 3
Solution Given, 2 sin
= tan −1
5
7
2

3
3
 3
LHS = 2 sin −1 = sin −1 2 ×
1−   
 5 
5
5


3
4
24


 
= sin −1 2 × ×
= sin −1  

 25
5 5 
Question 3. 2 sin −1


24



25
= tan −1 

2
 1 −  24 
 25 

[Q2 sin −1 y = sin −1(2y 1 − y 2 )]

y 
Q sin −1 y = tan −1

1 − y 2 

NCERT Class XII Mathematics Solutions


24


25 
24 
−1  24
25
= tan 
×
= tan −1
= RHS.
 = tan 

 7 
25
7
576


 1−

625 
8
3
77
Question 4. sin −1 + sin −1 = tan −1
17
5
36
−1
59
Hence proved.
For adding two sin−1 functions, we use the relation
sin−1 x + sin−1 y = sin−1( x 1 − y 2 + y 1 − x 2 ).
8
3
77
+ sin −1 = tan −1
17
5
36
2
2
8
3
 3
 8
 3
 8
LHS = sin −1   + sin −1   = sin −1 
1−   +
1−   
 5
 17
 5
 17 
5
17

Solution Given, sin −1
[∴sin −1 x + sin −1 y = sin −1(x 1 − y 2 + y 1 − x 2 ]
 8 4 3 15
 77
= sin −1 
× + ×  = sin −1  
 17 5 5 17
 85


77



85
= tan −1 

2
 1 −  77 
 85 

77
 77 85 
= tan −1
×
= tan −1
= RHS.
85 36 
36

x 
Q sin −1 x = tan −1

1 − x 2 

Hence proved.
4
12
33
+ cos −1
= cos −1 .
5
13
65
33
−1 4
−1 12
Solution Given, cos
+ cos
= cos −1
5
13
65
 4
 12
LHS = cos −1   + cos −1  
 5
 13
Question 5. cos −1
2
2
 4 12
 4
 12
= cos −1  .
− 1−   1−   
 5 13
 5
 13 


[Qcos −1 x + cos −1 y = cos −1(xy − 1 − x 2 1 − y 2 ]
2
2
 48
5
 48 3
 3  5 
= cos −1 
−      = cos −1 
− × 
 65
 5  13 
 65 5 13


3
 48
 33
= cos −1 
−  = cos −1   = RHS.
 65 13
 65 
Hence proved.
60
www.arihantbooks.com
Question 6. cos −1
12
3
56
+ sin −1 = sin −1
13
5
65
In LHS there are the different inverse trigonometric function. So, we
12
convert cos −1
to sin−1 θ form and then use the formula
13
sin−1 x + sin−1 y = sin−1( x 1 − y 2 + y 1 − x 2 ).
12
3
56
+ sin −1 = sin −1
13
5
65
12
12
cos −1
= θ ⇒ cos θ =
13
13
Solution Given, cos −1
Let
∴
 12
sin θ = 1 −  
 13
2
[Qsin θ = 1 − cos 2 θ ]
5
25
5
⇒ θ = sin −1
=
13
169 13
3
−1 12
−1 3
−1 5
Now,
LHS = cos
+ sin
= sin
+ sin −1
13
5
13
5
2
2
 5
3
 5
 3
1−   +
1−   
= sin −1 
 13
 13 
 5
5


⇒
sin θ =
[Qsin −1 x + sin −1 y = sin −1(x 1 − y 2 + y 1 − x 2 )]
 5 16 3 144 
 56
 5 4 3 12
= sin −1 
+
× + ×  = sin −1   = RHS
 = sin −1 
 65 
 13 5 5 13
 13 25 5 169 
Hence proved.
Question 7. tan −1
63
5
3
= sin −1
+ cos −1 .
16
13
5
5
Here, we put sin−1   = x and then convert x in terms of tan−1. Similarly
 13
3
put cos −1   = y and then convert y in terms of tan−1 and then apply the
 5
 a+ b
formula tan−1 a + tan−1 b = tan−1 
.
 1 − ab 
63
5
3
= sin −1
+ cos −1
16
13
5
−1 5
−1 3
RHS = sin
+ cos
13
5
5
−1  5 
sin   = x ⇒ sin x =
 13
13
Solution Given, tan −1
Let
2
Q
 5
1 − sin 2 x = cos 2 x ⇒ 1 −   = cos 2 x
 13
NCERT Class XII Mathematics Solutions
61
169 − 25
144
12
= cos x
= cos 2 x ⇒
= cos 2 x ⇒
169
169
13
5
sin x
5
5
13
∴
tan x =
=
⇒ tan x =
⇒ x = tan −1
12 12
cos x
12
13
3
3
Let
cos −1 = y ⇒ cos y =
5
5
2
 3
Q
1 − cos 2 y = sin 2 y ⇒ 1 −   = sin 2 y
 5
25 − 9
16
4
⇒
= sin y
= sin 2 y ⇒
= sin 2 y ⇒
25
25
5
4
sin y
4
4
tan y =
⇒ y = tan −1
⇒ tan y = 5 =
∴
3
3
cos y
3
5
63
∴ Given equation becomes tan −1
=x+y
16
63
 4
 5
⇒ tan −1
= tan −1   + tan −1  
 3
 12
16
⇒
4 
 5
+
5
4






RHS = tan −1   + tan −1   = tan −1  12 3 
∴
5 4
 12
 3
× 
1 −

12 3 

−1
−1
−1  x + y  
Q tan x + tan y = tan  1 − xy  


 15 + 48 
 12 × 3 
36 
63 
−1  63
= LHS
= tan −1 
×
= tan −1
 = tan 

16 
×
−
12
3
20
 36 36 − 20 


 12 × 3 
LHS = RHS.
Hence proved.
∴
1
1
1
1
π
Question 8. tan −1 + tan −1 + tan −1 + tan −1 =
5
7
3
8 4
We do not add all tan−1 x function one at a time, so we make a pair and
 x + y
use the formula tan−1 x + tan−1 y = tan−1 
.
 1 − xy 
1
1
+ tan −1 +
5
7
1
−1 1
+ tan
 +
5
7
Solution Given, tan −1

LHS =  tan −1

1
1 π
+ tan −1 =
3
8 4
1

−1 1
+ tan −1 
 tan

3
8
tan −1
62
www.arihantbooks.com
 1 1 
 1 1 
+
+




= tan −1  5 7  + tan −1  3 8 
 1 − 1 × 1
 1 − 1 × 1


5 7
3 8

−1
−1
−1  x + y  
Q tan x + tan y = tan  1 − xy  


 8+ 3
 7+ 5 




 6
 11
= tan −1  35  + tan −1  24  = tan −1   + tan −1  
35
1
24
1
−
−


 23
17




 24 
 35 
11 
 6
+


−1  6 × 23 + 11 × 17
17
23
= tan 
 + tan 

6
11
 17 × 23 − 6 × 11
1 −
× 

17 23 
−1
π
 325
−1
= tan −1 
 = tan 1 = = RHS.
 325
4
Question 9. tan −1 x =
Hence proved.
1 − x
1
cos −1 
 , x ∈[ 0, 1].
1 + x
2
1
2
1 − x
 , x ∈[0, 1]
1 + x
Solution Given, tan −1 x = cos −1 
LHS = tan −1 x =
1
1 − ( x )2
1
× (2 tan −1 x ) = × cos −1
2
2
1 + ( x )2
2 

−1
−1  1 − x 

Q2 tan x = cos 
1 + x2 

=
1 − x
1
cos −1 
 = RHS.
1 + x
2
Hence proved.
 1 + sin x + 1 − sin x  x
 π
 = , x ∈  0, 
 4
 1 + sin x − 1 − sin x  2
Question 10. cot −1 
To remove the square root sign, we use the identity 1 = sin2
x
x
the relation sin x = 2 sin ⋅ cos and also use the identities
2
2
x
x
+ cos 2 and
2
2
( a + b) 2 = a2 + b2 + 2ab and ( a − b) 2 = a2 + b2 − 2ab.
 1 + sin x + 1 − sin x  x
 π
 = , x ∈ 0, 
 4
 1 + sin x − 1 − sin x  2
Solution Given, cot −1 
NCERT Class XII Mathematics Solutions
 1 + sin x + 1 − sin x 

LHS = cot −1 
 1 + sin x − 1 − sin x 
63
…(i)
Now, we can write
1 + sin x = sin 2
x
x
x
x
x
x

+ cos 2 + 2 sin cos =  sin + cos 

2
2
2
2
2
2
2
x
x
x
x

Q sin 2 + cos 2 = 1 and sin x = 2 sin cos

2
2
2
2 
x
x
⇒
1 + sin x = sin + cos
2
2
x
x
Similarly, we can get 1 − sin x = cos − sin
2
2
On substituting above two values in Eq. (i) , we get
x
x
x
x
x


 sin + cos + cos − sin 
 2 cos 
−
1
−1
2
2
2
2
2  = cot 
LHS = cot 
 sin x + cos x + sin x − cos x 
 2 sin x 


2
2
2
2
2
x x

= cot −1 cot  = = RHS

2 2
Alternative method
 1 + sin x + 1 − sin x 
LHS = cot −1 

 1 + sin x − 1 − sin x 
 1 + sin x + 1 − sin x
1 + sin x + 1 − sin x 
= cot −1 
×

+
−
−
+ sin x + 1 − sin x 
sin
sin
1
x
1
x
1

(by rationalyzing denominator)
 ( 1 + sin x + 1 − sin x )2 
= cot −1 

2
( 1 + sin x) − ( 1 − sin x )2 
1 + sin x + 1 − sin x + 2 1 − sin 2 x 
= cot −1 

1 + sin x − 1 + sin x


= cot −1
2 + 2 cos x 
[Q cos 2 x + sin 2 x = 1 ⇒ cos x = 1 − sin 2 x ]
 2 sin x 
= cot −1

 2 cos 2 x /2
1 + cos x 
= cot −1 

 sin x 
2 sin x /2 cos x /2 
x
x

2 x
and sin x = 2 sin cos 
Q1 + cos x = 2 cos

2
2
2
64
www.arihantbooks.com
x x

= = RHS
= cot −1 cot

2  2
x
x
 π
π 
Note If x ∈ 0,  , then 1 − sin x = cos − sin and if x ∈  , π  , then
 2
2 
2
2
x
x
1 − sin x = sin − cos
2
2
 1 + x − 1 − x π 1
 = − cos −1 x
+
+
−
1
1
x
x

 4 2
Question 11. tan −1 
To remove the square root sign, firstly substitute x = cosθ then use the
relation 1 + cos θ = 2cos 2( θ/ 2) and 1 − cos θ = 2sin2( θ/ 2).
Solution Let x = cos θ so that cos −1 x = θ
 1 + cos θ − 1 − cos θ 
 1+ x − 1− x

 = tan −1 
tan −1 
 1 + cos θ + 1 − cos θ 
 1+ x + 1− x
θ
θ

 2 cos − 2 sin  
−1
2
2  Q1 + cos θ = 2 cos 2 θ and 1 − cos θ = 2 sin 2 θ 
= tan 
θ
2
2
 2 cos + 2 sin θ  

2
2
θ
θ
θ


 cos − sin 
 1 − tan 
−1
−1
2
2
2
= tan 
 = tan 
 cos θ + sin θ 
 1 + tan θ 


2
2
2
∴
(inside the bracket divide numerator and denominator by cos θ /2)

tan A − tan B 

 π θ 
= tan −1  tan  −  
Q tan (A − B) = 1 + tan A tan B 
 4 2 



π θ π 1
−1
Hence proved.
= − = − cos x
4 2 4 2
Question 12.
 2 2
9π 9
 1 9
− sin −1   = sin −1 

 3 4
8
4
 3 
9
1
Firstly, we shift the term sin−1   from LHS to RHS and use the relation
 3
4
sin−1 x + sin−1 y = sin−1( x 1 − y 2 + y 1 − x 2 ), to get the required result.
Solution Given,
2 2
9π 9
 1 9
− sin −1   = sin −1 

 3 4
8
4
 3 
⇒
2 2
9π 9
 1 9
= sin −1   + sin −1 

 3 4
8
4
 3 
⇒
2 2 
9 π 9  −1  1 
= sin   + sin −1 

 3
8
4
 3 
NCERT Class XII Mathematics Solutions
RHS =
=
65

9  −1  1 
−1  2 2 

sin   + sin 
4
3
 3 
2


2 
2 2
9  −1  1
2 2
 1 
1− 
1 −   
sin 
 +
 3 
4
3
 3 

 3


[Qsin −1 x + sin −1 y = sin −1(x 1 − y 2 + y 1 − x 2 )]
=
9  −1  1 1 2 2 2 2  
×

sin  × +
4
3
3 
3 3
9 π 9π
9  −1  1 8   9
= LHS. Hence proved.
sin  +   = sin −1 (1) = × =
 9 9  4
4 
4 2
8
Alternative method
9π 9
 1 9  π
 1  9 
 1 
LHS =
− sin −1   =  − sin −1    = cos −1   
 3 4  2
 3  4 
 3 
8
4
=
π

Q sin −1 x + cos −1 x =

2 
2
=
9
 1
sin −1 1 −  
 3
4
=
9
2 2
1 9
= RHS.
sin −1 1 − = sin −1
4
3
9 4
[Qcos −1 x = sin −1 1 − x 2 ]
Hence proved.
Direction (Q. Nos. 13 to 17) Solve the following equations for x :
Question 13. 2 tan −1 (cos x) = tan −1 ( 2 cosec x)
Solution Given, 2 tan −1(cos x) = tan −1(2 cosec x)
⇒
⇒
⇒


−1
−1  2 x 

Q2 tan x = tan 
2 


x
1
−


2 cos x
2
2 cos x
−1  2 cos x 
−1  2 
=
⇒
=2
tan 
 = tan 
 ⇒
 sin 2 x 
 sin x 
sin x
sin 2 x sin x
π
π
cot x = 1 ⇒ cot x = cot
⇒ x=
4
4
 2 cos x 
tan −1 
 = tan −1(2 cosec x)
 1 − cos 2 x 
Question 14. tan −1
1− x 1
= tan −1 x, ( x > 0).
1+ x 2
Multiply both sides by 2 and then on LHS use the relation
 2x 
 and solve it and equate both sides to get the
2 tan−1 x = tan−1 
2
1 − x 
value of x.
66
www.arihantbooks.com
Solution Given,
1 − x
−1
or 2 tan −1 
 = tan x
1 + x
1 − x 1
−1
tan −1 
 = tan x
1 + x 2
⇒
 1 − x 
 2
 
1 + x 
tan −1 
= tan −1 x
2



1 −  1 − x  
 1 + x  



−1
−1  2 y 

Q2 tan y = tan 
2 
1 − y  


2(1 − x)

(1 + x)
 = tan −1 x
2
2
x) − (1 − x) 

(1 + x)2

⇒


tan −1 
 (1 +


⇒
 2(1 − x)(1 + x) 
= tan −1 x
tan −1 
2
2
1
x
1
x
+
−
−
(
)
(
)


⇒


2(1 − x 2)
−1
tan −1  2
 = tan x
2
2
2
1
x
2
x
1
x
2
x
+
+
−
−
+


⇒
2(1 − x 2) 
−1
tan −1 
 = tan x
4
x


⇒
1 − x2
=x
2x
 (1 − x 2)
⇒ tan −1 
 = tan −1 x
 2x 
⇒ 1 − x 2 = 2x 2 ⇒ 1 = 3x 2 ⇒ x 2 =
1
1
⇒x = ±
3
3
[Qx > 0 given, so we do not take x = −
1
3
Question 15. sin (tan −1 x), |x| < 1, is equal to
x
1
1
(a)
(b)
(c)
2
2
1− x
1− x
1 + x2
1
]
3
x=
⇒

Solution (d) sin (tan −1 x) = sin sin −1
=

x
1 + x2
.
(d)
x
1 + x2


x
−1
−1
 Q tan x = sin
2
1+ x  
1+ x 
x
2
Hence, the correct option is (d).
π
2
Question 16. sin −1 (1 − x) − 2 sin −1 x = , then x is equal to
(a) 0,
1
2
(b) 1,
1
2
(c) zero
(d)
1
2
NCERT Class XII Mathematics Solutions
67
(i) Shift the term 2sin−1 x from LHS to RHS and use the identity
π
cos −1 a = − sin−1 a.
2
(ii) Take cosine on both sides and use cos( − x ) = cos x and
cos 2x = 1 − 2sin2 x.
Solution (c) Given, sin −1(1 − x) − 2 sin −1 x =
⇒
⇒
⇒
⇒
⇒
⇒
⇒
π
2
π
− sin −1(1 − x) ⇒ − 2 sin −1 x = cos −1(1 − x)
2
π

Q sin −1(1 − x) + cos −1(1 − x) =

2 
−1
[multiplying both sides by cos x]
cos (− 2 sin x) = 1 − x
−2 sin −1 x =
cos (2 sin −1 x) = 1 − x
[1 − 2 sin (sin
2
−1
[Qcos (− x) = + cos x]
x)] = 1 − x
1 − 2 [sin(sin −1 x)]
2
[Qcos 2x = 1 − 2 sin 2 x]
=1− x
[Qsin 2 x = (sin x)2]
1 − 2x = 1 − x ⇒ 2x − x = 0
2
2
x(2x − 1) = 0 ⇒ x = 0 or 2x − 1 = 0 ⇒ x = 0 or
1
2
1
does not satisfy the given equation, so x = 0.
2
Hence, the correct option is (c).
Alternate method
π
Given, sin −1(1 − x) − 2 sin −1 x =
2
Let x = sin θ ⇒ θ = sin −1 x, then
π
π
sin −1(1 − sin θ) − 2 θ =
⇒ sin −1(1 − sin θ ) = + 2 θ
2
2


π


π
1 − sin θ = sin  + 2 θ ⇒ 1 − sin θ = cos 2 θ Q sin  + θ = cos θ 
⇒

2
2



But x =
⇒
⇒
Either
⇒
⇒
1 − sin θ = 1 − 2 sin 2 θ
[Qcos 2 θ = 1 − 2 sin 2 θ]
2 sin 2 θ − sin θ = 0 ⇒ sin θ (2 sin θ − 1) = 0
sin θ = 0 or 2 sin θ − 1 = 0
[Qsin θ = x]
x = 0 or 2x − 1 = 0
1
x = 0 or x =
2
1
does not satisfy the given equation. So, x = 0.
2
To check your answer
Put x = 0 in the given equation,
π
π
π
⇒
sin −1 (1 − 0) − 2 sin −1 0 =
−2 ×0 =
⇒
∴
2
2
2
But x =
π π
=
2 2
68
www.arihantbooks.com
1
in the given equation,
2
1
1 π

∴
sin −1 1 −  − 2 sin −1 =

2
2 2
π
π π
1
1 π
⇒
sin −1 − 2 sin −1 =
⇒ −2 × =
6
6 2
2
2 2
1
π − 2π π
−π π
⇒
= ⇒
≠ , so x = is not possible.
2
6
2
6
2
Note While solving the trigonometric equation, sometimes it may have
some extra roots, so please take careful about it.
Put x =
 x
 x − y
Question 17. tan −1   − tan −1 
 is equal to
 y
 x + y
−3 π
4

x
 − 1
−1  x 
−1  x − y 
−1  x 
−1  y

Solution (c) tan   − tan 
 = tan   − tan
 y
 x + y
 y
 x + 1



y


 x 
 x − y 
x
= tan −1   − tan −1 − tan −1 1 Q tan −1 x − tan −1 y = tan −1 

 1 + xy  
 y 
y
 
π
= tan −1 1 = . Hence, the correct option is (c).
4
(a)
π
2
(b)
π
3
(c)
π
4
(d)
Selected NCERT Exemplar Problems
5π 
13 π 
−1 
.
 + cos  cos


6
6 
5π 
13π 

−1 
tan −1  tan

 + cos cos


6
6 
Question 1. Find the value of tan −1  tan
Solution
π 



 π π 
= tan −1 tan  +   + cos −1 cos 2π +  



2 3 
6 


π
π


= tan −1 − cot
+ cos −1 cos 


3 
6
π

[Qtan  + θ = − tan θ and cos(2π + θ) = cos θ]
2

π

 π π 

= tan −1 − tan  −   + cos −1 cos



2
3
6 





π

Q tan  2 − θ = cot θ 


NCERT Class XII Mathematics Solutions
69
π


 π 

 π  π
+ cos −1 cos    = tan −1 tan  −   +
= tan −1 − tan

 6 
 6  6
6 


[Q tan (− θ) = − tan θ]
π π
= − + =0
6 6
 − 3 π 
 + .
 2  6

Question 2. Evaluate cos cos −1 

Firstly applying the relation cos −1( − x ) = π − cos −1 x and then simplify it.

 − 3
π

 3
π
Solution cos cos −1 
 +  = cos π − cos −1 
 + 
2

 6
 2  6


(Qcos −1(− x) = π − cos −1 x)
π π

= cos π − +
= cos π = − 1

6 6 
Question 3. Find the value of
 1 
 −π  
−1  1 
−1 
tan −1  −
 + cot 
 + tan sin   

 3
3
  2 
1
 −π  
−1  1 
−1 
 + tan sin   
 + cot 
 3
3
  2 
π
 1
−1  1 
−1 
[Q sin (− θ) = − sin θ]
= tan −1  −
 + cot 
 + tan  − sin 


 3
2
3


Solution tan −1  −
π
−π π

+ − tan −1  sin 

6
3
2
π
π π
π
= − tan −1(1) = − = −
6
6 4
12
=
[Qtan −1(− x) = − tan −1 x]
Question 4. Find the real solutions of the equation
tan −1 x( x + 1) + sin −1 x2 + x + 1 =
π
.
2
Here, we do not add two different inverse trigonometric function. So, we
use the domains of tan−1 x and sin−1 x and simplify them.
π
2
π
2
x + x +1=
2
Solution. tan −1 x(x + 1) + sin −1 x 2 + x + 1 =
⇒
tan −1 x 2 + x + sin −1
This equation holds, if
x 2 + x ≥ 0 and 0 ≤ x 2 + x + 1 ≤ 1
Now,
x 2 + x ≥ 0 and 0 ≤ x 2 + x + 1 ≤ 1
70
www.arihantbooks.com
⇒
x 2 + x ≥ 0 and x 2 + x + 1 ≤ 1
⇒
x + x ≥ 0 and x + x ≤ 0
2
⇒
[Qx 2 + x + 1 > 0 for all x]
2
x 2 + x = 0 ⇒ x = 0, − 1
Clearly, these two values satisfy the given equation. Hence, x = 0, − 1 are
the solutions of the given equation.
Question 5. Find the value of the expression
Solution
1

sin  2 tan −1  + cos(tan −1 2 2 ).

3
1

Let E = sin 2 tan −1  + cos (tan −1 2 2)

3
Now, 2 tan
and
−1


 2× 1 
1
3  = tan −1
= tan −1 
2

3
1
  
1 −   
 3 

3
−1 3
−1
4
= tan
= sin
2
4
 3
1+  
 4
 3
 
3
= sin −1  4  = sin −1
5
5
 
 4
tan −1(2 2) = cos −1
2
3
8
9

−1
−1 2 θ 
Q2 tan θ = tan

1 − θ2 




x

Q tan −1 x = sin −1 

2

 1 + x  

 1
= cos −1  
 3
1 + (2 2)
1
2

Q tan −1 x = cos −1


∴

1  


 1 + x2  


1

E = sin 2 tan −1  + cos (tan −1 2 2)

3
3
1 3 1 9 + 5 14



= sin  sin −1  + cos cos −1  = + =
=



5
3 5 3
15
15
3
Question 6. Solve the following equation : cos (tan −1 x) = sin  cot −1 

4
3
Solution cos (tan −1 x) = sin cot −1 

4

= sin  tan −1  = sin

3

 sin −1


4


3 +4 
4
2
2

Q tan −1 x = sin −1




x


 1 + x2  


NCERT Class XII Mathematics Solutions
4 4
 =
5 5
4
x) = cos (− tan −1 x) =
5
71

= sin  sin −1

⇒
cos (tan −1
[Q cos (x) = cos (− x)]
 4
tan −1 x = − tan −1 x = cos −1  
 5
⇒
tan −1 x = − tan −1 x = tan −1
⇒
⇒
x=
Question 7.
3
4

Qcos −1 x = tan −1


 1 − x2  



x  


3 −3
,
4 4
 1 + x2 + 1 − x2  π 1
 = + cos −1 x2 .
Prove that tan −1 
 1 + x2 − 1 − x2  4 2


Substitute x 2 by cos 2θ to remove the square root sign and then use the
relation 1 + cos 2θ = 2cos 2 θ and 1 − cos 2 θ = 2 sin2 θ.
 1 + x2
tan −1 
 1 + x 2
 1 + x2 +
LHS = tan −1 
 1 + x2 −

Solution Given,
+ 1 − x2  π 1
 = + cos −1 x 2
2
− 1 − x  4 2
1 − x 2 
1 − x 2 
x 2 = cos 2θ
Let
 1 + cos 2 θ + 1 − cos 2 θ 
 2 cos θ + 2 sin θ 
 = tan −1 
∴ LHS = tan −1 

 2 cos θ − 2 sin θ 
 1 + cos 2 θ − 1 − cos 2 θ 
[Q1 + cos 2θ = 2 cos 2 θ and 1 − cos 2θ = 2 sin 2 θ]
 cos θ + sin θ 
= tan −1 

 cos θ − sin θ 
[divide inside the bracket by cos θ]


π
 tan + tan θ 


+
tan
1
θ
−1
4
 = tan −1 tan  π + θ 
= tan −1 
 = tan 

 
4
π
 1 − tan θ 




 1 −  tan  (tan θ)


4
π
π 1
+ θ = + cos −1 x 2
4
4 2
∴ LHS = RHS.
=

tan θ + tan φ 
Q tan (θ + φ) = 1 − tan θ tan φ 


1

2
−1 2 
Q x = cos 2 θ = ⇒ θ = cos x


2
Hence proved.
72
www.arihantbooks.com
Question 8. Find the simplified form of cos −1  cos x +
3
5
x∈
4

sin x , where

5
 − 3π π 
, .
 4
4 
The angle of cos −1 be converted into a standard formula cos ( a − b) by
3
4
substituting
= cos θ and
= sin θ and then use the identity
5
5
−1
cos [cos ( a − b)] = a − b.
4
3

E = cos −1  cos x + sin x
5

5
3
…(i)
Let
cos θ =
5
4
Then,
…(ii)
sin θ =
5
On dividing Eq. (ii) by Eq. (i), we get
4
4
…(iii)
tan θ =
⇒ θ = tan −1
3
3
4
3

∴
E = cos −1  cos x + sin x = cos −1(cos θ cos x + sin θ sin x)
5

5
[using Eqs. (i) and (ii)]
[Qcos(a − b) = cos a cos b + sin a sin b]
= cos −1 [cos(θ − x)]
4
[from Eq. (iii)]
= θ − x = tan −1 − x
3
Solution Let
Question 9. If a1 , a2 , a3 ,....,an is an arithmetic progression with common
difference d, then evaluate the following expression



d 
d
−1 
tan tan −1 

 + tan 



1
a
a
1
a
a
+
+
1 2
2 3





d
d
+ tan −1 
 + ...... + 
 .
1 + a3 a4 
1 + an−1 an  
(i) Since, a1, a2,......, an is an arithmetic progression with common
difference d.
Then, d = a2 − a1 = a3 − a2 = .......... = an − an −1
 x − y
−1
−1
(ii) Use the relation tan−1 
 = tan x − tan y and then simplify.
 1 + xy 
Solution Given, a1, a2, a3, a4,..., an is an arithmetic progression, then
d = a2 − a1 = a4 − a3 = ..... = an − an −1
NCERT Class XII Mathematics Solutions
∴
73

 d 
d 
−1 
tan tan −1 
 + tan 




+
1
a
a
1
a2a3 
+
1 2



d 
d
−1 
+ tan −1 
 + .... + tan 

 1 + a3a4 
 1 + an −1an  

 a − a1 
−1  a − a2 
= tan tan −1  2
 + tan  3

 1 + a1a2 
 1 + a2a3 

 a − a3 
−1  a − an −1  
+ tan −1  4
 + .... + tan  n

 1 + a3a4 
 1 + an −1an  
= tan[tan −1 a2 − tan −1 a1 + tan −1 a3 − tan −1 a2 + tan −1 a4
− tan −1 a3 + ..... + tan −1 an − tan −1 an −1]

−1  x − y 
−1
−1 
Q tan 1 + xy  = tan x − tan y 





an − a1
−1
−1
−1  an − a1  
= tan[tan an − tan a1] = tan tan 
 =
 1 + ana1   1 + a1an
