cot_1 (1+x/1-x)=1/2 cot _1 (1/x)
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Answer:
Given that,
y
=
sin
4
(
cot
−
1
√
1
−
x
1
+
x
)
,
we need
d
y
d
x
.
We substitute
x
=
cos
2
θ
,
so that,
−
1
<
x
<
1
.
Note that, in order to make
√
1
−
x
1
+
x
meaningful, we must have
−
1
<
x
<
1
,
which justifies our substitution :
x
=
cos
2
θ
.
∴
y
=
sin
4
(
cot
−
1
√
1
−
x
1
+
x
)
,
=
sin
4
(
cot
−
1
√
1
−
cos
2
θ
1
+
cos
2
θ
)
,
=
sin
4
⎛
⎝
cot
−
1
√
2
sin
2
θ
2
cos
2
θ
⎞
⎠
,
=
sin
4
(
cot
−
1
(
tan
θ
)
)
,
=
sin
4
(
cot
−
1
{
cot
(
π
2
−
θ
)
}
)
,
=
sin
4
(
π
2
−
θ
)
,
=
{
sin
(
π
2
−
θ
)
}
4
,
=
cos
4
θ
,
=
(
cos
2
θ
)
2
,
=
{
1
+
cos
2
θ
2
}
2
.
⇒
y
=
1
4
(
1
+
x
)
2
...
...
...
...
...
...
...
...
...
...
...
...
...
.
[
∵
,
cos
2
θ
=
x
]
.
∴
d
y
d
x
=
1
4
⋅
d
d
x
(
1
+
x
)
2
,
=
1
4
⋅
2
(
1
+
x
)
⋅
d
d
x
(
1
+
x
)
...
...
...
...
.
.
[
∵
,
the Chain Rule],
⇒
d
y
d
x
=
1
2
(
1
+
x
)
.
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