Then sin–1 x = θ We have
sin–1 (
2)
2
x1
−x
 = sin–1 (
)
2
2sin
1
sin
θ
−
θ
= sin–1 (2sinθ cosθ) = sin–1 (sin2θ) = 2θ
= 2 sin–1 x
(ii)
Take x = cos θ, then proceeding as above, we get, sin–1 (
2)
2
x1
−x
= 2 cos–1 x
Example 4 Express 
1
cos
tan
1
sin
x
x
−  
 
 
 
 
 
−
, 
23
2
π
π
−
<
x<
 in the simplest form Solution We write
2
2
1
–1
2
2
cos
sin
cos
2
2
tan
tan
1
sin
cos
sin
2sin
cos
2
2
2
2
x
x
x
x
x
x
x
x
−


−



 =




−



+
−


=
–1
2
cos
sin
cos
sin
2
2
2
2
tan
cos
sin
2
2
x
x
x
x
x
x





+
−














−








=
–1 cos
sin
2
2
tan
cos
sin
2
2
x
x
x
x


+






−


 
–1 1
tan 2
tan
1
tan 2
x
x


+

=




−

=
tan–1
tan 4
2
4
2
x
x


π
π


+
=
+








Rationalised 2023-24
INVERSE TRIGONOMETRIC FUNCTIONS     29
Example 5 Write 
–1
12
cot
1
x



−


,  x  > 1 in the simplest form Solution Let x = sec θ, then 
2
1
x − = 
sec2
1
θ − =tan
θ
Therefore, 
–1
12
cot
1
x −
= cot–1 (cot θ) = θ = sec–1 x, which is the simplest form