CHAPTER 2
INVERSE TRIGONOMETRIC FUNCTIONS
IMPORTANT POINTS

sin–1 x, cos–1 x, ... etc., are angles.

 
If sin  x and    ,  then  = sin–1x etc.
 2 2 


Function
Domain
Range
(Principal Value Branch)
sin–1x
[–1, 1]
  
  2 , 2 
cos–1x
[–1, 1]
[0, ]
tan–1x
R
  
  , 
2 2
cot–1x
R
(0, )
sec–1x
R – (–1, 1)
0,   
cosec –1x
R – (–1, 1)
    
  2 , 2   0


2
sin–1 (sin x) = x  x     ,  
 2 2 
cos–1 (cos x) = x  x  [0, ] etc.

sin (sin–1x) = x  x  [–1, 1]
cos (cos–1x) = x  x  [–1, 1] etc.
17
XII – Maths

 1
sin1x  cosec –1   x  1, 1
x
tan–1x = cot–1 (1/x)  x > 0
sec–1x = cos–1 (1/x),  |x|  1

sin–1(–x) = – sin–1x  x  [–1, 1]
tan–1(–x) = – tan–1x  x  R
cosec–1(–x) = – cosec–1x  |x|  1

cos–1(–x) =  – cos–1x  ×  [–1, 1]
cot–1(–x) =  – cot–1x  x  – R
sec–1(–x) =  – sec–1x  |x|  1

sin1 x  cos 1 x 
tan–1 x  cot –1 x 

, x   1, 1
2

 x R
2
sec –1 x  cosec –1x 

 x 1
2

x  y
tan1 x  tan1 y  tan1 
;
 1  xy 
xy  1.

x  y
tan1 x  tan1 y  tan 1 
;
 1  xy 
xy  1.

 2x 
2 tan–1 x  tan –1 
, x 1
 1  x 2 
 2x 
2 tan1 x  sin1 
, x  1,
 1  x 2 
 1 x 2 
2 tan1 x  cos 1 
, x  0.
 1  x 2 
XII – Maths
18
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)
1.
Write the principal value of
–1
–
3 2
(ii)
cos
–1
1 

 –

3
(iv)
cosec–1 (– 2).
–1
 1 

.
3
(vi)
sec–1 (– 2).
1
 3
1  1
1

  cos    tan  1 3 
2
2
(i)
sin
(iii)
tan
(v)
cot
(vii)
sin
–1

3 2 .
2.
What is value of the following functions (using principal value).
(i)
tan
–1
–1  2 
 1 

 – sec 
.
3
3
(iv)
cosec
tan–1 (1) – cot–1 (–1).
(v)
tan–1 (1) + cot–1 (1) + sin–1 (1).
(vi)
sin
(viii)
4 

 sin
.
5 
cosec
–1
–1  3 
 1
.
 –  – cos 
2
2 
sin
(iii)
–1
–1
(ii)
(vii)
tan
–1

–1
5 

 tan
.
6 
2   sec
–1

2.
3 

 cosec
.
4 
SHORT ANSWER TYPE QUESTIONS (4 MARKS)
3.
Show that tan
–1
 1  cos x 

 1  cos x –
19
1  cos x 

x
 .
 
1 – cos x 
4
2
x  [0, ]
XII – Maths
4.
Prove
1
tan
5.
Prove
6.
Prove
cot
–1

1  1  cos x 
 cos x 
 

  cot 
1  sin x
1  cos x 
4
tan
–1
x   0,  2  .
2
2 

x
a  x
–1 x
–1


.
 sin
 cos 




a
a
 a2 – x 2 

–1 8  
–1 
–1 8  
–1  300 


 2 tan  cos
   tan  2 tan  sin
   tan 
.
17 
17 
161 


1
7.
Prove
tan
8.
Solve
cot
–1
2

1 x 

2

1 x 
2x  cot
–1


1
1 2
 
 cos x .
2 
4
2
1 x 
1 x
3x 

2
.
4
9.
 m
 m  n 
Prove that tan1    tan1 
 , m, n  0
n
 m  n  4
10.
2 
1
x y
 2x  1
1  1  y 

cos
Prove that tan  sin1 
 1  y 2    1  xy
2


2
2
1

x



11.
 x 2  1 1
 2x  2

Solve for x, cos1  2   tan1 
 x  1 2
 1 x 2 
3
12.
Prove that tan1
13.
Solve for
14.
 1
 1
 32 
Prove that 2 tan–1    tan –1    tan –1  
 4
 5
 43 
XII – Maths
x,
1
1
1
1 
 tan1  tan1  tan1 
3
5
7
8 4
tan cos 1 x   sin tan1 2 ; x  0
20
1
 3 
tan  cos –1 
 11 
2

15.
Evaluate
16.
 a cos x  b sin x 
 a
 tan –1    x
Prove that tan–1 
 b cos x  a sin x 
b
17.
Prove that

 1 
cot tan1 x  tan1     cos 1 1  2 x 2   cos 1 2 x 2  1  , x  0
 x

18.
 a b 
 b c 
 c a 
 tan –1 
 tan –1 
 0 where a, b,
Prove that tan–1 
 1  ab 
 1  bc 
 1  ca 
c > 0
19.
Solve for x, 2 tan–1(cos x) = tan–1 (2 cosec x)
20.
Express
21.
If tan–1a + tan–1b + tan–1c =  then

sin–1 x 1 x  x 1 x 2

in simplest form.
prove that a + b + c = abc
22.
If sin–1x > cos–1x, then x belongs to which interval?
ANSWERS
1.
2.
(ii)

6
(iii)
–
6

3
(vi)
2
3
(vii)

.
6
(i)
0
(ii)

3
(iii)

(v)

(vi)

5
(vii)
(i)
–
(v)

3
21
(iv)
–
6

2
(iv)

2
–
6
(viii)

.
4
XII – Maths
8.
1
11.
13.
5
3
19.
x 
22.
 1

, 1

2 
21.
Hint:
15.

.
4
Let
20
tan–1 a = 
tan–1 b = 
tan–1 c = 
then given,       

      
take tangent on both sides,
tan (  ) = tan   
XII – Maths
22
tan

 2 3
12
11  3
3  11
sin–1 x – sin–1 x.