Prove the identity arcsin((x-1)/(x+1))=2arctan-

One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in x=0
x
=
0
.

Here's another way. Look at
π2−2arctanx−−√=2(π4−arctanx−−√)=2(arctan1−arctanx−−√).
π
2
−
2
arctan
⁡
x
=
2
(
π
4
−
arctan
⁡
x
)
=
2
(
arctan
⁡
1
−
arctan
⁡
x
)
.
Now remember the identity for the difference of two arctangents:
arctanu−arctanv=arctanu−v1+uv.
arctan
⁡
u
−
arctan
⁡
v
=
arctan
⁡
u
−
v
1
+
u
v
.
(This follows from the usual identity for the tangent of a sum.) The left side above becomes
2arctan1−x−−√1+x−−√.
2
arctan
⁡
1
−
x
1
+
x
.
The double-angle formula for the sine says sin(2u)=2sinucosu
sin
⁡
(
2
u
)
=
2
sin
⁡
u
cos
⁡
u
. Apply that:
sin(2arctan1−x−−√1+x−−√)=2sin(arctan1−x−−√1+x−−√)cos(arctan1−x−−√1+x−−√)
sin
⁡
(
2
arctan
⁡
1
−
x
1
+
x
)
=
2
sin
⁡
(
arctan
⁡
1
−
x
1
+
x
)
cos
⁡
(
arctan
⁡
1
−
x
1
+
x
)
Now remember that sin(arctanu)=u1+u2−−−−−√
sin
⁡
(
arctan
⁡
u
)
=
u
1
+
u
2
and cos(arctanu)=11+u2−−−−−√
cos
⁡
(
arctan
⁡
u
)
=
1
1
+
u
2
Then use algebra:
2⋅(1−x√1+x√)1+(1−x√1+x√)2−−−−−−−−−−√⋅11+(1−x√1+x√)2−−−−−−−−−−√=1−x1+x.