Properties of Inverse Trigonometric Functions

# Properties of Inverse Trigonometric Functions | Mathematics (Maths) Class 12 - JEE PDF Download

``` Page 1

D. PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
P-1 (i) y = sin (sin
–1
x) = x (ii) y = cos (cos
–1
x) = x
x ? [–1, 1], y ? [–1, 1] x ? [–1, 1], y ? [–1, 1]
45º
0
y = x
1
x
y
–1
–1
1
45º
0
y = x
1
x
y
–1
–1
1
(iii) y = tan (tan
–1
x) = x (iv) y = cot (cot
–1
x) = x
x ? R, y ? R x ? R, y ? R
45º
0
y = x
x
y
45º
0
y = x
x
y
(v) y = cosec (cosec
–1
x) = x (vi) y = sec (sec
–1
x) = x
| x | ? 1, | y | ? 1 | x | ? 1 ; | y | ? 1
0
y = x
1
x
y
–1
–1
1
y = x
0
y = x
1
x
y
–1
–1
1
y = x
Page 2

D. PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
P-1 (i) y = sin (sin
–1
x) = x (ii) y = cos (cos
–1
x) = x
x ? [–1, 1], y ? [–1, 1] x ? [–1, 1], y ? [–1, 1]
45º
0
y = x
1
x
y
–1
–1
1
45º
0
y = x
1
x
y
–1
–1
1
(iii) y = tan (tan
–1
x) = x (iv) y = cot (cot
–1
x) = x
x ? R, y ? R x ? R, y ? R
45º
0
y = x
x
y
45º
0
y = x
x
y
(v) y = cosec (cosec
–1
x) = x (vi) y = sec (sec
–1
x) = x
| x | ? 1, | y | ? 1 | x | ? 1 ; | y | ? 1
0
y = x
1
x
y
–1
–1
1
y = x
0
y = x
1
x
y
–1
–1
1
y = x
(vii) y = sin
–1
(sin x), x ? R, y ? ?
?
?
?
?
? ? ?
?
2
,
2
, is periodic function with period 2 ? and it is an odd function
sin
–1
(sin x) =
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
? ?
?
?
?
? ? ? ? ? ? ? ?
x
2
, x
2
x
2
, x
2
x , x

/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(viii) y = cos
–1
(cos x), x ? R, y ? [0, ?], is periodic function with period 2 ? and it is an even function
cos
–1
(cos x) =
x , x 0
x , 0 x
2 x , x 2
x 2 , 2 x 3
? ? ? ? ? ?
? ? ? ?
?
? ? ? ? ? ?
?
? ? ? ? ? ?
?
y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
(ix) y = tan
–1
(tan x), x ? R – ?
(2n 1) , n I
2
? ? ?
? ?
? ?
? ?
;  y ?
,
2 2
? ? ? ?
?
? ?
? ?
is periodic function with period ? and it
is an odd function
tan
–1
(tan x) =
3
x ; x
2 2
x ; x
2 2
3
x ; x
2 2
? ? ?
? ? ? ? ? ?
?
?
? ? ?
? ? ?
?
?
? ?
?
? ? ? ?
?
?

? ?2
? 2
0
y=x
y=x+ ?
2
3 ?
?
? ?
2
?
?
2
? ?
y=x– ?
2
3 ?
2
?
2
?
?
x
y
(x) y = cot
–1
(cot x), x ? R – {n ?, n ? I}, y ? [0, ?], is periodic function with period ? and it is nei-
ther an even nor odd function
cot
–1
(cot x) =
?
?
?
?
?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
2 x ; x
x 0 ; x
0 x ; x
? ?2 ? 2 0
y=x
y=x+ ?
? ? ?
?
y=x– ?
x
y
y=x+2 ?
(xi) y = cosec
–1
(cosec x), x ? R – {n ?, n ? I}, y ?
?
?
?
?
?
? ?
? ?
?
?
?
?
? ?
?
2
, 0 0 ,
2
is periodic function with period
2 ? and it is an odd function
Page 3

D. PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
P-1 (i) y = sin (sin
–1
x) = x (ii) y = cos (cos
–1
x) = x
x ? [–1, 1], y ? [–1, 1] x ? [–1, 1], y ? [–1, 1]
45º
0
y = x
1
x
y
–1
–1
1
45º
0
y = x
1
x
y
–1
–1
1
(iii) y = tan (tan
–1
x) = x (iv) y = cot (cot
–1
x) = x
x ? R, y ? R x ? R, y ? R
45º
0
y = x
x
y
45º
0
y = x
x
y
(v) y = cosec (cosec
–1
x) = x (vi) y = sec (sec
–1
x) = x
| x | ? 1, | y | ? 1 | x | ? 1 ; | y | ? 1
0
y = x
1
x
y
–1
–1
1
y = x
0
y = x
1
x
y
–1
–1
1
y = x
(vii) y = sin
–1
(sin x), x ? R, y ? ?
?
?
?
?
? ? ?
?
2
,
2
, is periodic function with period 2 ? and it is an odd function
sin
–1
(sin x) =
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
? ?
?
?
?
? ? ? ? ? ? ? ?
x
2
, x
2
x
2
, x
2
x , x

/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(viii) y = cos
–1
(cos x), x ? R, y ? [0, ?], is periodic function with period 2 ? and it is an even function
cos
–1
(cos x) =
x , x 0
x , 0 x
2 x , x 2
x 2 , 2 x 3
? ? ? ? ? ?
? ? ? ?
?
? ? ? ? ? ?
?
? ? ? ? ? ?
?
y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
(ix) y = tan
–1
(tan x), x ? R – ?
(2n 1) , n I
2
? ? ?
? ?
? ?
? ?
;  y ?
,
2 2
? ? ? ?
?
? ?
? ?
is periodic function with period ? and it
is an odd function
tan
–1
(tan x) =
3
x ; x
2 2
x ; x
2 2
3
x ; x
2 2
? ? ?
? ? ? ? ? ?
?
?
? ? ?
? ? ?
?
?
? ?
?
? ? ? ?
?
?

? ?2
? 2
0
y=x
y=x+ ?
2
3 ?
?
? ?
2
?
?
2
? ?
y=x– ?
2
3 ?
2
?
2
?
?
x
y
(x) y = cot
–1
(cot x), x ? R – {n ?, n ? I}, y ? [0, ?], is periodic function with period ? and it is nei-
ther an even nor odd function
cot
–1
(cot x) =
?
?
?
?
?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
2 x ; x
x 0 ; x
0 x ; x
? ?2 ? 2 0
y=x
y=x+ ?
? ? ?
?
y=x– ?
x
y
y=x+2 ?
(xi) y = cosec
–1
(cosec x), x ? R – {n ?, n ? I}, y ?
?
?
?
?
?
? ?
? ?
?
?
?
?
? ?
?
2
, 0 0 ,
2
is periodic function with period
2 ? and it is an odd function
/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(xii) y = sec
–1
(sec x), x ? R – ?
?
?
?
?
?
?
?
?
? I n ,
2
) 1 n 2 (
, y ?
0, ,
2 2
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
is periodic function with period 2 ?
and it is an even function

y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
Ex.3 Evaluate following
(i) sin(cos
–1
3/5) (ii) cos(tan
–1
3/4) (iii) sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
2
1
sin
2
1
Sol. (i) Let cos
–1
3/5 = ? then cos ? = 3/5  ?  sin ? = 4/5 ? ?? ? ?  sin(cos
–1
3/5) = sin ? = 4/5
(ii) Let tan
–1
3/4 = ? then tan ? = 3/4  ?  cos ? = 4/5 ? ? ? ? ?  cos(tan
–1
3/4) = cos ? = 4/5
(iii) sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
2
1
sin
2
1
= sin ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
6 2
= sin
3
2 ?
=
2
3
Ex.4 Define the function,  f(x) = cos
?1
(cos x) ? sin
?1
(sin x)  in  [0, 2

?]  and  find the area bounded by the
graph of the function and the  x

?

axis.
Sol. cos
?1
(cos x)  =
x
x
x
x 2
0
2 ?
?
? ? ?
? ?
? ?
?
?
?
; sin
?1
(sin x)  =
x
x
x
x
x
x
?
?
?
?
?
?
?
?
?
? ?
? ?
? ?
?
?
?
?
?
?
2
0
2
2
2
3
2
3
2
Hence    f

(x) =
? ?
? ?
? ?
? ?
0 0
2
4 2 2
2
2
3
2
3
2
i f x
x i f x
i f x
x i f x
?
? ?
?
? ?
?
?
?
?
?
?
?
?
,
,
,
,
?
?
?
?
? ?
? ?
? ?
Area =
3
2 2
? ?
?
?
?
?
?
?
? ?
?
2
=  ?
2
Ex.5 Let  y = sin
–1
(sin 8) – tan
–1
(tan 10) + cos
–1
(cos 12) – sec
–1
(sec 9) + cot
–1
(cot 6) – cosec
–1
(cosec 7).
If  y  simplifies to  a ? + b  then find ( a – b).
Page 4

D. PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
P-1 (i) y = sin (sin
–1
x) = x (ii) y = cos (cos
–1
x) = x
x ? [–1, 1], y ? [–1, 1] x ? [–1, 1], y ? [–1, 1]
45º
0
y = x
1
x
y
–1
–1
1
45º
0
y = x
1
x
y
–1
–1
1
(iii) y = tan (tan
–1
x) = x (iv) y = cot (cot
–1
x) = x
x ? R, y ? R x ? R, y ? R
45º
0
y = x
x
y
45º
0
y = x
x
y
(v) y = cosec (cosec
–1
x) = x (vi) y = sec (sec
–1
x) = x
| x | ? 1, | y | ? 1 | x | ? 1 ; | y | ? 1
0
y = x
1
x
y
–1
–1
1
y = x
0
y = x
1
x
y
–1
–1
1
y = x
(vii) y = sin
–1
(sin x), x ? R, y ? ?
?
?
?
?
? ? ?
?
2
,
2
, is periodic function with period 2 ? and it is an odd function
sin
–1
(sin x) =
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
? ?
?
?
?
? ? ? ? ? ? ? ?
x
2
, x
2
x
2
, x
2
x , x

/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(viii) y = cos
–1
(cos x), x ? R, y ? [0, ?], is periodic function with period 2 ? and it is an even function
cos
–1
(cos x) =
x , x 0
x , 0 x
2 x , x 2
x 2 , 2 x 3
? ? ? ? ? ?
? ? ? ?
?
? ? ? ? ? ?
?
? ? ? ? ? ?
?
y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
(ix) y = tan
–1
(tan x), x ? R – ?
(2n 1) , n I
2
? ? ?
? ?
? ?
? ?
;  y ?
,
2 2
? ? ? ?
?
? ?
? ?
is periodic function with period ? and it
is an odd function
tan
–1
(tan x) =
3
x ; x
2 2
x ; x
2 2
3
x ; x
2 2
? ? ?
? ? ? ? ? ?
?
?
? ? ?
? ? ?
?
?
? ?
?
? ? ? ?
?
?

? ?2
? 2
0
y=x
y=x+ ?
2
3 ?
?
? ?
2
?
?
2
? ?
y=x– ?
2
3 ?
2
?
2
?
?
x
y
(x) y = cot
–1
(cot x), x ? R – {n ?, n ? I}, y ? [0, ?], is periodic function with period ? and it is nei-
ther an even nor odd function
cot
–1
(cot x) =
?
?
?
?
?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
2 x ; x
x 0 ; x
0 x ; x
? ?2 ? 2 0
y=x
y=x+ ?
? ? ?
?
y=x– ?
x
y
y=x+2 ?
(xi) y = cosec
–1
(cosec x), x ? R – {n ?, n ? I}, y ?
?
?
?
?
?
? ?
? ?
?
?
?
?
? ?
?
2
, 0 0 ,
2
is periodic function with period
2 ? and it is an odd function
/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(xii) y = sec
–1
(sec x), x ? R – ?
?
?
?
?
?
?
?
?
? I n ,
2
) 1 n 2 (
, y ?
0, ,
2 2
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
is periodic function with period 2 ?
and it is an even function

y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
Ex.3 Evaluate following
(i) sin(cos
–1
3/5) (ii) cos(tan
–1
3/4) (iii) sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
2
1
sin
2
1
Sol. (i) Let cos
–1
3/5 = ? then cos ? = 3/5  ?  sin ? = 4/5 ? ?? ? ?  sin(cos
–1
3/5) = sin ? = 4/5
(ii) Let tan
–1
3/4 = ? then tan ? = 3/4  ?  cos ? = 4/5 ? ? ? ? ?  cos(tan
–1
3/4) = cos ? = 4/5
(iii) sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
2
1
sin
2
1
= sin ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
6 2
= sin
3
2 ?
=
2
3
Ex.4 Define the function,  f(x) = cos
?1
(cos x) ? sin
?1
(sin x)  in  [0, 2

?]  and  find the area bounded by the
graph of the function and the  x

?

axis.
Sol. cos
?1
(cos x)  =
x
x
x
x 2
0
2 ?
?
? ? ?
? ?
? ?
?
?
?
; sin
?1
(sin x)  =
x
x
x
x
x
x
?
?
?
?
?
?
?
?
?
? ?
? ?
? ?
?
?
?
?
?
?
2
0
2
2
2
3
2
3
2
Hence    f

(x) =
? ?
? ?
? ?
? ?
0 0
2
4 2 2
2
2
3
2
3
2
i f x
x i f x
i f x
x i f x
?
? ?
?
? ?
?
?
?
?
?
?
?
?
,
,
,
,
?
?
?
?
? ?
? ?
? ?
Area =
3
2 2
? ?
?
?
?
?
?
?
? ?
?
2
=  ?
2
Ex.5 Let  y = sin
–1
(sin 8) – tan
–1
(tan 10) + cos
–1
(cos 12) – sec
–1
(sec 9) + cot
–1
(cot 6) – cosec
–1
(cosec 7).
If  y  simplifies to  a ? + b  then find ( a – b).
Sol. sin
–1
(sin 8) =
? ? ) 8 3 s i n( s i n
1
? ?
?
= 3 ? – 8
tan
–1
(tan 10) =
? ? ) 3 10 t a n( t a n
1
? ?
?
= 10 – 3 ?
cos
–1
(cos 12) =
? ? ) 1 2 4 c o s( c o s
1
? ?
?
= 4 ? – 12
sec
–1
(sec 9) =
? ? ) 2 9 s e c ( s e c
1
? ?
?
= 9 – 2 ?
cot
–1
(cot 6) =
? ? ) 6 c ot ( c ot
1
? ?
?
= 6 – ?
cosec
–1
(cosec 7) =
? ? ) 2 7 ( c o s e c c o s e c
1
? ?
?
= 7 – 2 ?
y = (3 ? – 8) + (3 ? ?– 10) + (4 ? – 12) + (2 ? – 9) + (– ? + 6 ) + (2 ? ?– 7) = 13 ? – 40
? a = 13  and  b = – 40 ? a – b = 13 – (– 40) = 53
P-2 (i) cosec
–1
x

=  sin
–1
x
1
; |x| ? ?1     (ii) sec
–1
x  = cos
–1
x
1
; |x| ? 1
(iii) cot
–1
x =
?
?
?
?
?
?
?
? ? ?
?
0 x ,
x
1
tan
0 x ,
x
1
tan
1 –
1 –
P-3  (i)  sin
–1
(–x) = – sin
–1
x  ;  – 1 ? x ? 1 (ii) cosec
–1
(–x) = – cosec
–1
x  ;  |x| ? 1
(iii) tan
–1
(–x) = – tan
–1
x  ;  x ? R (iv) cot
–1
(–x) = ? ?– cot
–1
x  ;  x ? R
(v) cos
–1
(–x) = ? ?– cos
–1
x  ;  – 1 ? x ? 1 (vi) sec
–1
(–x) = ? – sec
–1
x  ;  |x| ? 1
P-4  (i) sin
–1
x + cos
–1
x =
2
?
;  |x| ? 1 (ii) tan
–1
x + cot
–1
x  =
2
?
;  x ? R
(iii) sec
–1
x + cosec
–1
x  =
2
?
;  |x| ? ?1
Ex.6 Find the value of sin
–1
(– 2 / 3 ) + cos
–1
(cos (7 ?/6).
Sol. sin
–1
(– 2 / 3 ) = – sin
–1
( 2 / 3 ) = – ?/3 and cos
–1
(cos (7 ?/6) = cos
–1
cos (2 ? – 5 ?/6) = cos
–1
cos(5 ?/6) = 5 ?/6
hence sin
–1
(– 2 / 3 ) + cos
–1
(cos (7 ?/6) = –
3
?
+
6
5 ?
=
2
?
Ex.7 Prove that,  sin
–1

1
3

+  sin
–1

1
3 1 1

+  sin
–1

3
1 1

=

?
2
.
Sol. tan
–1

1
2 2

+

tan
–1

1
7 2

+

tan
–1

3
2
= tan
–1

1
2 2
1
7 2
1
1
2 8
?
?

+

tan
–1

3
2
=  tan
–1

9 2
2 7

+

tan
–1

3
2
=  tan
–1

2
3

+

tan
–1

3
2
=  cot
–1

3
2

+

tan
–1

3
2
=
?
2
Ex.8 Find the value of  sin
?1

73
3

+  cos
?1

146
11

+  cot
?1
3 .
Sol. tan
?1

3
8
+  tan
?1

11
5

+  cot
?1
3 =  tan
?1

?
?
?
?
?
?
?
?
?
?
?
?
11
5
8
3
11
5
8
3
. 1
+ cot
?1
3 =  tan
?1
(1)

+

?
6
=
?
4
+

?
6
=

5
12
?
Ex.9 If tan
–1
x + tan
–1
y + tan
–1
z =
2
3 ?
then prove that xy  + yz + zx = 1
Page 5

D. PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
P-1 (i) y = sin (sin
–1
x) = x (ii) y = cos (cos
–1
x) = x
x ? [–1, 1], y ? [–1, 1] x ? [–1, 1], y ? [–1, 1]
45º
0
y = x
1
x
y
–1
–1
1
45º
0
y = x
1
x
y
–1
–1
1
(iii) y = tan (tan
–1
x) = x (iv) y = cot (cot
–1
x) = x
x ? R, y ? R x ? R, y ? R
45º
0
y = x
x
y
45º
0
y = x
x
y
(v) y = cosec (cosec
–1
x) = x (vi) y = sec (sec
–1
x) = x
| x | ? 1, | y | ? 1 | x | ? 1 ; | y | ? 1
0
y = x
1
x
y
–1
–1
1
y = x
0
y = x
1
x
y
–1
–1
1
y = x
(vii) y = sin
–1
(sin x), x ? R, y ? ?
?
?
?
?
? ? ?
?
2
,
2
, is periodic function with period 2 ? and it is an odd function
sin
–1
(sin x) =
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
? ?
?
?
?
? ? ? ? ? ? ? ?
x
2
, x
2
x
2
, x
2
x , x

/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(viii) y = cos
–1
(cos x), x ? R, y ? [0, ?], is periodic function with period 2 ? and it is an even function
cos
–1
(cos x) =
x , x 0
x , 0 x
2 x , x 2
x 2 , 2 x 3
? ? ? ? ? ?
? ? ? ?
?
? ? ? ? ? ?
?
? ? ? ? ? ?
?
y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
(ix) y = tan
–1
(tan x), x ? R – ?
(2n 1) , n I
2
? ? ?
? ?
? ?
? ?
;  y ?
,
2 2
? ? ? ?
?
? ?
? ?
is periodic function with period ? and it
is an odd function
tan
–1
(tan x) =
3
x ; x
2 2
x ; x
2 2
3
x ; x
2 2
? ? ?
? ? ? ? ? ?
?
?
? ? ?
? ? ?
?
?
? ?
?
? ? ? ?
?
?

? ?2
? 2
0
y=x
y=x+ ?
2
3 ?
?
? ?
2
?
?
2
? ?
y=x– ?
2
3 ?
2
?
2
?
?
x
y
(x) y = cot
–1
(cot x), x ? R – {n ?, n ? I}, y ? [0, ?], is periodic function with period ? and it is nei-
ther an even nor odd function
cot
–1
(cot x) =
?
?
?
?
?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
2 x ; x
x 0 ; x
0 x ; x
? ?2 ? 2 0
y=x
y=x+ ?
? ? ?
?
y=x– ?
x
y
y=x+2 ?
(xi) y = cosec
–1
(cosec x), x ? R – {n ?, n ? I}, y ?
?
?
?
?
?
? ?
? ?
?
?
?
?
? ?
?
2
, 0 0 ,
2
is periodic function with period
2 ? and it is an odd function
/2
y
0
?
?
? ?
–  /2
?
y=–( +x) ?
2
3 ?
?
? ?2
x
2
?
?
2
?
2
3 ?
y=2 +x ?
y=x
? 2
y=x–2 ?
y=  –x ?
(xii) y = sec
–1
(sec x), x ? R – ?
?
?
?
?
?
?
?
?
? I n ,
2
) 1 n 2 (
, y ?
0, ,
2 2
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
is periodic function with period 2 ?
and it is an even function

y=x+2 ?
y
?
?
2 ?
0 ? ?
–3 ?/2 –2 ?
– ?/2 ?/2 3 ?/2
y=–x
y=x
y=2 –x ?
2
?
Ex.3 Evaluate following
(i) sin(cos
–1
3/5) (ii) cos(tan
–1
3/4) (iii) sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
2
1
sin
2
1
Sol. (i) Let cos
–1
3/5 = ? then cos ? = 3/5  ?  sin ? = 4/5 ? ?? ? ?  sin(cos
–1
3/5) = sin ? = 4/5
(ii) Let tan
–1
3/4 = ? then tan ? = 3/4  ?  cos ? = 4/5 ? ? ? ? ?  cos(tan
–1
3/4) = cos ? = 4/5
(iii) sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
2
1
sin
2
1
= sin ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
6 2
= sin
3
2 ?
=
2
3
Ex.4 Define the function,  f(x) = cos
?1
(cos x) ? sin
?1
(sin x)  in  [0, 2

?]  and  find the area bounded by the
graph of the function and the  x

?

axis.
Sol. cos
?1
(cos x)  =
x
x
x
x 2
0
2 ?
?
? ? ?
? ?
? ?
?
?
?
; sin
?1
(sin x)  =
x
x
x
x
x
x
?
?
?
?
?
?
?
?
?
? ?
? ?
? ?
?
?
?
?
?
?
2
0
2
2
2
3
2
3
2
Hence    f

(x) =
? ?
? ?
? ?
? ?
0 0
2
4 2 2
2
2
3
2
3
2
i f x
x i f x
i f x
x i f x
?
? ?
?
? ?
?
?
?
?
?
?
?
?
,
,
,
,
?
?
?
?
? ?
? ?
? ?
Area =
3
2 2
? ?
?
?
?
?
?
?
? ?
?
2
=  ?
2
Ex.5 Let  y = sin
–1
(sin 8) – tan
–1
(tan 10) + cos
–1
(cos 12) – sec
–1
(sec 9) + cot
–1
(cot 6) – cosec
–1
(cosec 7).
If  y  simplifies to  a ? + b  then find ( a – b).
Sol. sin
–1
(sin 8) =
? ? ) 8 3 s i n( s i n
1
? ?
?
= 3 ? – 8
tan
–1
(tan 10) =
? ? ) 3 10 t a n( t a n
1
? ?
?
= 10 – 3 ?
cos
–1
(cos 12) =
? ? ) 1 2 4 c o s( c o s
1
? ?
?
= 4 ? – 12
sec
–1
(sec 9) =
? ? ) 2 9 s e c ( s e c
1
? ?
?
= 9 – 2 ?
cot
–1
(cot 6) =
? ? ) 6 c ot ( c ot
1
? ?
?
= 6 – ?
cosec
–1
(cosec 7) =
? ? ) 2 7 ( c o s e c c o s e c
1
? ?
?
= 7 – 2 ?
y = (3 ? – 8) + (3 ? ?– 10) + (4 ? – 12) + (2 ? – 9) + (– ? + 6 ) + (2 ? ?– 7) = 13 ? – 40
? a = 13  and  b = – 40 ? a – b = 13 – (– 40) = 53
P-2 (i) cosec
–1
x

=  sin
–1
x
1
; |x| ? ?1     (ii) sec
–1
x  = cos
–1
x
1
; |x| ? 1
(iii) cot
–1
x =
?
?
?
?
?
?
?
? ? ?
?
0 x ,
x
1
tan
0 x ,
x
1
tan
1 –
1 –
P-3  (i)  sin
–1
(–x) = – sin
–1
x  ;  – 1 ? x ? 1 (ii) cosec
–1
(–x) = – cosec
–1
x  ;  |x| ? 1
(iii) tan
–1
(–x) = – tan
–1
x  ;  x ? R (iv) cot
–1
(–x) = ? ?– cot
–1
x  ;  x ? R
(v) cos
–1
(–x) = ? ?– cos
–1
x  ;  – 1 ? x ? 1 (vi) sec
–1
(–x) = ? – sec
–1
x  ;  |x| ? 1
P-4  (i) sin
–1
x + cos
–1
x =
2
?
;  |x| ? 1 (ii) tan
–1
x + cot
–1
x  =
2
?
;  x ? R
(iii) sec
–1
x + cosec
–1
x  =
2
?
;  |x| ? ?1
Ex.6 Find the value of sin
–1
(– 2 / 3 ) + cos
–1
(cos (7 ?/6).
Sol. sin
–1
(– 2 / 3 ) = – sin
–1
( 2 / 3 ) = – ?/3 and cos
–1
(cos (7 ?/6) = cos
–1
cos (2 ? – 5 ?/6) = cos
–1
cos(5 ?/6) = 5 ?/6
hence sin
–1
(– 2 / 3 ) + cos
–1
(cos (7 ?/6) = –
3
?
+
6
5 ?
=
2
?
Ex.7 Prove that,  sin
–1

1
3

+  sin
–1

1
3 1 1

+  sin
–1

3
1 1

=

?
2
.
Sol. tan
–1

1
2 2

+

tan
–1

1
7 2

+

tan
–1

3
2
= tan
–1

1
2 2
1
7 2
1
1
2 8
?
?

+

tan
–1

3
2
=  tan
–1

9 2
2 7

+

tan
–1

3
2
=  tan
–1

2
3

+

tan
–1

3
2
=  cot
–1

3
2

+

tan
–1

3
2
=
?
2
Ex.8 Find the value of  sin
?1

73
3

+  cos
?1

146
11

+  cot
?1
3 .
Sol. tan
?1

3
8
+  tan
?1

11
5

+  cot
?1
3 =  tan
?1

?
?
?
?
?
?
?
?
?
?
?
?
11
5
8
3
11
5
8
3
. 1
+ cot
?1
3 =  tan
?1
(1)

+

?
6
=
?
4
+

?
6
=

5
12
?
Ex.9 If tan
–1
x + tan
–1
y + tan
–1
z =
2
3 ?
then prove that xy  + yz + zx = 1
Sol. Since tan
–1
x + tan
–1
y + tan
–1
z =
2
3 ?
?   tan
–1
x + tan
–1
y =
2
3 ?
– tan
–1
z
?   tan (tan
–1
x + tan
–1
y) = tan
?
?
?
?
?
?
?
?
?
z tan
2
3
1
? ?
xy 1
y x
?
?
= cot (tan
–1
z) ...(1)
Case (I) : If z > 0 then tan
–1
z = cot
–1

?
?
?
?
?
?
z
1
?  cot (tan
–1
z) = cot
?
?
?
?
?
?
? ? ?
?
z
1
cot
1
=
z
1
...(2)
Case (II) : If z < 0 then tan
–1
z = – ? + cot
–1
?
?
?
?
?
?
z
1
? cot (tan
–1
z) = cot
?
?
?
?
?
?
? ? ?
?
z
1
cot
1
= cot
?
?
?
?
?
?
?
z
1
cot
1
=
z
1
...(3)
From (1), (2) and (3) we get
z
1
xy 1
y x
?
?
?
or zx + yz = 1 – xy or xy + yz + zx = 1.
Ex.10 If cos
–1
x/2 + cos
–1
y/3 = ?, prove that 9x
2
+ 12 xy cos ? + 4y
2
= 36sin
2
?
Sol. Let cos
–1
x/2 = ? ? and  cos
–1
y/3 = ? ? cos ? = x/2 and cos ? =  y/3.
Given, ? ? ? ? ? ? ? ? ? cos ( ? ? ? ? ?) = cos ?
or cos ? cos ? – sin ? sin ? = ? or
9
y
1
4
x
1
3
y
.
2
x
2 2
? ? ? = cos ?
or
6
y 9 . x 4
6
xy
2 2
? ?
? = cos ? or (xy – 6cos ?)
2
= (4 – x
2
) (9 – y
2
)
or x
2
y
2
+ 36cos
2
? – 12xy cos ? = 36 – 9x
2
– 4y
2
+ x
2
y
2
or 9x
2
– 12y cos ? + 4y
2
= 36 (1 – cos
2
?) or 9x
2
– 12xycos ? + 4y
2
= 36sin
2
?.
Ex.11 If u = cos
–1

? 2 cos
– tan
–1

? 2 cos
, prove that sin u = tan
2
?
Sol. Given, u = tan
–1
? 2 cos
1
– tan
–1
? 2 cos
= tan
–1

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
2 cos .
2 cos
1
1
2 cos
2 cos
1
= tan
–1
?
?
?
?
?
?
?
? ?
2 cos 2
2 cos 1
= tan
–1

?
?
2 cos
sin
2
? tan u =
?
?
2 cos
sin
2
=
BC
AB
(say)
? 2 cos
C
B
A
cos
2
?
sin
2
?
u
Then AC =
? ? ? 2 cos sin
4
=
? ? ? ?
2 4
sin 2 1 sin
= cos
2
?
? sinu =
? ?
?
?
?
2
2
2
tan
cos
sin
AC
AB
.
Ex.12 Show that cos
–1

7
1
+ 2 cot
–1

3
1
=
4
5 ?
Sol. cot
–1

7
1
+ 2 cot
–1

3
1
=
2
?
– tan
=1

7
1
+ 2
?
?
?
?
?
?
?
?
?
3
1
tan
2
1
=
?
?
?
?
?
?
? ?
?
? ?
3
1
tan 2
7
1
tan
2
3
1 1
=
?
?
?
?
?
?
?
?
?
? ?
?
? ?
2
1 1
) 3 / 1 ( 1
3 / 1 . 2
tan 2
7
1
tan
2
3
?
?
?
?
?
?
? 1
3
1
?
```

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

## FAQs on Properties of Inverse Trigonometric Functions - Mathematics (Maths) Class 12 - JEE

 1. What are the properties of inverse trigonometric functions?
Ans. The properties of inverse trigonometric functions are as follows: - The domain of an inverse trigonometric function is the range of the corresponding trigonometric function. - The range of an inverse trigonometric function is the domain of the corresponding trigonometric function. - The inverse trigonometric functions are all one-to-one functions. - The inverse trigonometric functions can be used to find the angle measures when given the ratios of the sides of a right triangle. - The inverse trigonometric functions have a restricted range in order to ensure that they have unique values.
 2. How are inverse trigonometric functions denoted?
Ans. Inverse trigonometric functions are denoted using the prefix "arc" or "a" followed by the abbreviation of the corresponding trigonometric function. For example, the inverse sine function is denoted as "arcsin" or "asin", the inverse cosine function as "arccos" or "acos", and the inverse tangent function as "arctan" or "atan".
 3. What is the relationship between trigonometric functions and their inverses?
Ans. The relationship between trigonometric functions and their inverses is that they "undo" each other. When a trigonometric function is applied to an angle, the resulting value can be used as an input for the corresponding inverse trigonometric function, which will give back the original angle. For example, if we take the sine of an angle and then take the arcsine of the resulting value, we will obtain the original angle.
 4. How are inverse trigonometric functions used in real-life applications?
Ans. Inverse trigonometric functions are used in various real-life applications, such as: - Calculating the angle of elevation or depression in geometry and physics. - Solving problems related to navigation and surveying, such as determining distances and angles. - Analyzing waveforms in electrical engineering and signal processing. - Modeling and analyzing periodic phenomena in fields like physics, engineering, and biology. - Solving problems involving triangles and angles in trigonometry and geometry.
 5. What are the principal values of inverse trigonometric functions?
Ans. The principal values of inverse trigonometric functions are the values that lie within a certain range and are used to ensure the functions have unique values. The principal values of inverse trigonometric functions are generally defined as follows: - For arcsine (arcsin), the principal values lie between -π/2 and π/2, inclusive. - For arccosine (arccos), the principal values lie between 0 and π, inclusive. - For arctangent (arctan), the principal values lie between -π/2 and π/2, exclusive.

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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