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INVERSE TRIGONOMETRIC FUNCTIONS     47
Therefore,
–1
2
1
cot
x −1
= cot
–1
(cot θ) = θ = sec
–1
x, which is the simplest form.
Example 7 Prove that tan
–1
x +
–1
2
2
tan
1
x
−x
= tan
–1
3
2
3
1 3
x x
x
1
| |
3
x <
Solution Let x = tan θ. Then θ = tan
–1
x. We have
R.H.S. =
3
3
–1
–1
2
2
3
3tan
tan
tan
tan
1 3
1 3tan
x x
x
θ−
θ
=
θ
=tan
–1
(tan3θ) = 3θ = 3tan
–1
x = tan
–1
x + 2 tan
–1
x
=tan
–1
x + tan
–1
2
2
1
x
−x
= L.H.S. (Why?)
Example 8 Find the value of  cos (sec
–1
x + cosec
–1
x), | x | ≥ 1
Solution We have cos (sec
–1
x + cosec
–1
x) = cos
2
π
 
 
 
= 0
EXERCISE 2.2
Prove the following:
1. 3sin
–1
x = sin
–1
(3x – 4x
3
),
1 1
– ,
2 2
x
2. 3cos
–1
x = cos
–1
(4x
– 3x),
1
,1
2
x
3. tan
–1
1
1
2
7
1
tan
tan
11
24
2
+
=
4.
1
1
1
1
1
31
2tan
tan
tan
2
7
17
+
=
Write the following functions in the simplest form:
5.
2
1
1
1
tan
x
x
+
, x ≠ 0
6.
1
2
1
tan
1
x
, |x | > 1
7.
1
1 cos
tan
1 cos
x
x
+
, 0 < x <  π
8.
1
cos
sin
tan
cos
sin
x
x
x
x
+
, 0 < x< π
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48
MATHEMATICS
9.
1
2
2
tan
x
a
x
, |x | < a
10.
2
3
1
3
2
3
tan
3
a x x
a
ax
, a > 0;
3
3
< <
a
a
x
Find the values of each of the following:
11.
–1
–1
1
tan 2 cos 2sin
2
12. cot (tan
–1
a  + cot
–1
a)
13.
2
–1
–1
2
2
1
2
1
tan
sin
cos
2
1
1
x
y
x
y
+
+
+
, | x| < 1, y > 0 and xy < 1
14. If  sin
–1
–1
1
sin
cos
1
5
x
+
=
, then find the value of x
15. If
–1
–1
1
1
tan
tan
2
2 4
x
x
x
x
+
π
+
=
+
, then find the value of x
Find the values of each of the expressions in Exercises 16 to 18.
16.
–1
2
sin
sin
3
π
17.
–1
3
tan
tan
4
π
18.
–1
–1
3
3
tan sin
cot
5
2
+
19.
1
7
cos
cos
is equal to
6
π
(A)
7
6
π
(B)
5
6
π
(C)
3
π
(D)
6
π
20.
1
1
sin
sin ( )
3
2
π
is equal to
(A)
1
2
(B)
1
3
(C)
1
4
(D) 1
21.
1
1
tan
3 cot ( 3)
is equal to
(A) π
(B)
2
π
(C) 0
(D)
2 3
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INVERSE TRIGONOMETRIC FUNCTIONS     49
Miscellaneous Examples
Example 9 Find the value of
1
3
sin (sin
)
5
π
Solution We know that
1
sin (sin x)
x
x
=
. Therefore,
1
3
3
sin (sin
)
5
5
π
π
=
But
3
,
5
2 2
π
π π
∉ −
, which is the principal branch of sin
–1
x
However
3
3
2
sin ( ) sin(
) sin
5
5
5
π
π
π
=
π−
=
and
2
,
5
2 2
π
π π
∈ −
Therefore
1
1
3
2
2
sin (sin
) sin (sin
)
5
5
5
π
π
π
=
=
Example 10 Show that
1
1
1
3
8
84
sin
sin
cos
5
17
85
=
Solution Let
1
3
sin
5
x
=
and
1
8
sin
17
y
=
Therefore
3
sin
5
x=
and
8
sin
17
y=
Now
2
9
4
cos
1 sin
1
25
5
x
x
=
=
=
(Why?)
and
2
64
15
cos
1 sin
1
289 17
y
y
=
=
=
We have
cos (x–y) = cos x cos y + sin x sin y
=
4 15 3
8
84
5 17 5 17 85
×
+ ×
=
Therefore
1
84
cos
85
x y
 
− =
 
 
Hence
1
1
1
3
8
84
sin
sin
cos
5
17
85
=
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50
MATHEMATICS
Example 11 Show that
1
1
1
12
4
63
sin
cos
tan
13
5
16
+
+
Solution Let
1
1
1
12
4
63
sin
, cos
,tan
13
5
16
x
y
z
=
=
=
Then
12
4
63
sin
, cos
, tan
13
5
16
x
y
z
=
=
=
Therefore
5
3
12
3
cos
,sin
,tan
and tan
13
5
5
4
x
y
x
y
=
=
=
=
We have
tan
tan
tan(
)
1 tan tan
x
y
x y
x
y
+
+
=
12 3
63
5 4
12 3
16
1
5 4
+
=
=−
− ×
Hence
tan(
)
tan
x y
z
+
=−
i.e.,
tan (x + y) = tan (–z) or tan (x + y) = tan (π – z)
Therefore
x + y = –  z  or  x + y = π – z
Since
x, y and z are positive, x + y ≠ – z  (Why?)
Hence
x + y + z = π  or
–1
–1
–1
12
4
63
sin
cos
tan
13
5
16
+
+
Example 12 Simplify
–1
cos
sin
tan
cos
sin
a
x b
x
b
x a
x
+
, if
a
b
tan x > –1
Solution We have,
–1
cos
sin
tan
cos
sin
a
x b
x
b
x a
x
+
=
–1
cos
sin
cos
tan
cos
sin
cos
a
x b
x
b
x
b
x a
x
b
x
+
–1
tan
tan
1
tan
a
x
b
a
x
b
+
=
–1
–1
tan
tan (tan )
a
x
b
–1
tan
a
x
b
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INVERSE TRIGONOMETRIC FUNCTIONS     51
Example 13 Solve tan
–1
2x + tan
–1
3x =
4
π
Solution We have tan
–1
2x + tan
–1
3x =
4
π
or
–1
2
3
tan
1 2
3
x
x
x
x
+
×
=
4
π
i.e.
–1
2
5
tan
1 6
x
x
=
4
π
Therefore
2
5
1 6
x
− x
=
tan
1
4
π
=
or
6x
2
+ 5x – 1 = 0 i.e., (6x – 1) (x + 1) = 0
which gives
x =
1
6
or x = – 1.
Since x = – 1 does not satisfy the equation, as the L.H.S. of the equation becomes
negative,
1
6
x=
is the only solution of the given equation.
Miscellaneous Exercise on Chapter 2
Find the value of the following:
1.
–1
13
cos
cos
6
π
2.
–1
7
tan
tan
6
π
Prove that
3.
–1
–1
3
24
2sin
tan
5
7
=
4.
–1
–1
–1
8
3
77
sin
sin
tan
17
5
36
+
=
5.
–1
–1
–1
4
12
33
cos
cos
cos
5
13
65
+
=
6.
–1
–1
–1
12
3
56
cos
sin
sin
13
5
65
+
=
7.
–1
–1
–1
63
5
3
tan
sin
cos
16
13
5
=
+
8.
–1
1
1
1
1
1
1
1
tan
tan
tan
tan
5
7
3
8 4
π
+
+
+
=
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52
MATHEMATICS
Prove that
9.
–1
–1
1
1
tan
cos
2
1
x
x
x
=
+
, x ∈ [0, 1]
10.
–1
1 sin
1 sin
cot
2
1 sin
1 sin
x
x
x
x
x
+
+
=
+
0,
4
x
π
11.
–1
–1
1
1
1
tan
cos
4
2
1
1
x
x
x
x
x
+ −
π
= −
+ +
1
1
2
x
≤ ≤
[Hint: Put x = cos 2θ]
12.
1
1
9
9
1 9
2 2
sin
sin
8
4
3 4
3
π
=
Solve the following equations:
13. 2tan
–1
(cos  x) = tan
–1
(2 cosec x) 14.
–1
–1
1
1
tan
tan
,(
0)
1
2
x
x x
x
=
>
+
15. sin (tan
–1
x), | x| < 1 is equal to
(A)
2
1
x
−x
(B)
2
1
1−x
(C)
2
1
1+x
+
(D)
2
1
x
+x
16. sin
–1
(1 – x) – 2 sin
–1
x =
2
π
, then x is equal to
(A) 0,
1
2
(B) 1,
1
2
(C) 0
(D)
1
2
17.
1
1
tan
tan
x
x y
y
x y
 
 
+
 
is equal to
(A)
2
π
(B)
3
π
(C)
4
π
(D)
3
4
− π
INVERSE TRIGONOMETRIC FUNCTIONS     53
Summary
Î
The domains and ranges (principal value branches) of inverse trigonometric
functions are given in the following table:
Functions
Domain
Range
(Principal Value  Branches)
y = sin
–1
x
[–1, 1]
,
2 2
−π π
y = cos
–1
x
[–1, 1]
[0, π]
y = cosec
–1
x
R – (–1,1)
,
2 2
−π π
– {0}
y = sec
–1
x
R – (–1, 1)
[0, π] –
{ }
2
π
y = tan
–1
x
R
,
2 2
π π
y = cot
–1
x
R
(0, π)
Î
sin
–1
x should not be confused  with (sin
x)
–1
. In  fact (sin x)
–1
1
sin x
and
similarly for other trigonometric functions.
Î
The value of an inverse trigonometric functions which lies in its principal
value  branch is  called  the principal  value of  that  inverse  trigonometric
functions.
For suitable values of domain, we have
Î
y = sin
–1
x ⇒ x = sin y
Î
x = sin y  ⇒ y = sin
–1
x
Î
sin (sin
–1
x) = x
Î
sin
–1
(sin x) =  x
Î
sin
–1
1
x
= cosec
–1
x
Î
cos
–1
(–x)  = π  – cos
–1
x
Î
cos
–1
1
x
=  sec
–1
x
Î
cot
–1
(–x) = π  – cot
–1
x
Î
tan
–1
1
x
= cot
–1
x
Î
sec
–1
(–x) = π  – sec
–1
x
54
MATHEMATICS
Î
sin
–1
(–x) =  – sin
–1
x
Î
tan
–1
(–x) = – tan
–1
x
Î
tan
–1
x + cot
–1
x =
2
π
Î
cosec
–1
(–x) =  –  cosec
–1
x
Î
sin
–1
x + cos
–1
x =
2
π
Î
cosec
–1
x +  sec
–1
x =
2
π
Î
tan
–1
x + tan
–1
y = tan
–1
1
x y
xy
+
Î
2tan
–1
=  tan
–1
2
2
1
x
−x
Î
tan
–1
x –  tan
–1
y = tan
–1
1
x y
xy
+
Î
2tan
–1
x = sin
–1
2
2
1
x
+x
= cos
–1
2
2
1
1
x
x
+
Historical Note
The study  of trigonometry  was first  started  in  India. The  ancient  Indian
Mathematicians, Aryabhatta (476A.D.), Brahmagupta  (598 A.D.), Bhaskara I
(600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. All
this knowledge went from India to Arabia and then from there to Europe. The
Greeks had  also started the  study of  trigonometry  but  their approach was so
clumsy that when the Indian approach became known, it was immediately adopted
throughout the world.
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents one of
the  main contribution  of the  siddhantas  (Sanskrit  astronomical works)  to
mathematics.
Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions
for angles more  than  90°. A  sixteenth  century Malayalam work Yuktibhasa
contains a proof for the expansion of sin (A + B). Exact expression for sines or
cosines of 18°, 36°, 54°, 72°, etc., were given by Bhaskara II.
The symbols sin
–1
x, cos
–1
x, etc., for arc sin x, arc cos x, etc., were suggested
by  the  astronomer  Sir  John  F.W.  Hersehel  (1813) The  name  of  Thales
(about 600 B.C.) is invariably associated with height and distance problems. He
is credited with the determination of the height of a great pyramid in Egypt by
measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known